Extension and trace results for doubling metric measure spaces and their hyperbolic fillings
aa r X i v : . [ m a t h . M G ] A ug Extension and trace results for doubling metricmeasure spaces and their hyperbolic fillings
Anders Bj¨orn
Department of Mathematics, Link¨oping University, SE-581 83 Link¨oping, [email protected], ORCID : Jana Bj¨orn
Department of Mathematics, Link¨oping University, SE-581 83 Link¨oping, [email protected], ORCID : Nageswari Shanmugalingam
Department of Mathematical Sciences, University of Cincinnati, P.O. Box 210025,Cincinnati, OH 45221-0025, U.S.A. ; [email protected], ORCID : Abstract
In this paper we study connections between Besov spaces of functions on a com-pact metric space Z , equipped with a doubling measure, and the Newton–Sobolev space offunctions on a uniform domain X ε . This uniform domain is obtained as a uniformization ofa (Gromov) hyperbolic filling of Z . To do so, we construct a family of hyperbolic fillings inthe style of Bonk–Kleiner [9] and Bourdon–Pajot [13]. Then for each parameter β > µ β of the doubling measure ν on Z to X ε , and show that µ β is doubling andsupports a 1-Poincar´e inequality. We then show that for each θ with 0 < θ < p ≥ β = p (1 − θ ) log α such that the Besov space B θp,p ( Z ) is the trace spaceof the Newton–Sobolev space N ,p ( X ε , µ β ) when ε = log α . Finally, we exploit the tools ofpotential theory on X ε to obtain fine properties of functions in B θp,p ( Z ), such as their qua-sicontinuity and quasieverywhere existence of L q -Lebesgue points with q = s ν p/ ( s ν − pθ ),where s ν is a doubling dimension associated with the measure ν on Z . Applying this tocompact subsets of Euclidean spaces improves upon a result of Netrusov [42] in R n . Key words and phrases : Gromov hyperbolic space, hyperbolic filling, Poincar´e inequality,doubling measure, uniform space, uniformization, trace and extension, Besov space, Besovcapacity, Sobolev capacity, quasicontinuity, Lebesgue point.Mathematics Subject Classification (2020): Primary: 31E05, Secondary: 30L05, 53C23.
Contents ∂ ε X
125 Uniformizing a hyperbolic filling with parameter ε ≤ log α
176 Roughly similar equivalence 197 Trees 228 Geodesics in the hyperbolic filling 249 Measures, function spaces and capacities 29 Anders Bj¨orn, Jana Bj¨orn and Nageswari Shanmugalingam
10 Lifting doubling measures from Z to its hyperbolic filling X Z from the hyperbolic filling X Z to its hyperbolic filling X Z
1. Introduction
Much of the current trend in first-order analysis (such as Sobolev type spaces andpotential theory) on metric measure spaces assumes that the underlying space isat least locally compact, doubling and supports a Poincar´e type inequality, seefor example [2] and [27]. On doubling spaces that do not support any Poincar´einequality, there are other possible choices of function spaces that are, however,nonlocal. This means that the energy of functions in these spaces depends on theirglobal behavior and can be nonzero on subsets where the functions vanish or areconstant. Examples of such spaces include Besov, Triebel–Lizorkin and (fractional)Haj lasz–Sobolev spaces. As locality is a highly useful tool in potential theory andvariational problems, it is desirable to seek an alternative approach to studyingfunction spaces on nonsmooth metric spaces without Poincar´e inequalities.Using a construction termed hyperbolic filling , Bonk–Kleiner [9, Theorem 11.1]and Bourdon–Pajot [13] connected compact doubling metric spaces to Gromov hy-perbolic spaces.
Gromov hyperbolicity is a notion of negative curvature in thenonsmooth metric setting and, unlike Alexandrov curvature which covers all scales,it captures negative curvature at large scales by requiring that every point in ageodesic triangle is within a bounded distance from the other two sides. Gro-mov hyperbolicity has proven to be a highly useful tool in studying the conformalgeometry of hyperbolic groups (Bridson–Haefliger [14] and Gromov [22]) and inunderstanding uniform domains (Bonk–Heinonen–Koskela [8], Bonk–Schramm [12]and Herron–Shanmugalingam–Xie [28]). We refer the interested reader to [14] andBuyalo–Schroeder [16] for more on synthetic notions of curvature in the metricsetting and the hyperbolic filling technique, respectively.In the current paper we contribute to the study of nonlocal analysis on compactdoubling metric measure spaces by introducing measures to the above constructionof hyperbolic fillings and by subsequently linking the nonlocal Besov spaces oncompact doubling metric measure spaces to the Newtonian (Sobolev) spaces onuniformizations of their hyperbolic fillings, see Theorem 1.1.One of the main results of the paper is that every
Besov space B θp,p ( Z ) with0 < θ < p ≥ Theorem 1.1.
Let Z be a compact metric space equipped with a doubling measure ν , X be a hyperbolic filling of Z with parameters α, τ > , and X ε be the uniformizationof X with parameter ε = log α .Then for each parameter β > we can equip X ε with a measure µ β induced by ν so that µ β is doubling and supports a -Poincar´e inequality both on X ε and itscompletion X ε . Moreover, for each < θ < and ≤ p < ∞ , the Besov space B θp,p ( Z ) is the trace space of the Newtonian space N ,p ( X ε , µ β ) with β = εp (1 − θ ) . More precisely, Tr ◦ E is the identity map on B θp,p ( Z ), whereTr : N ,p ( X ε , µ β ) → B θp,p ( Z ) and E : B θp,p ( Z ) → N ,p ( X ε , µ β )are bounded linear trace and extension operators, respectively. xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 3 In fact, we construct Tr for each θ ≤ − β/εp (see Theorem 11.3), and E foreach θ ≥ − β/εp (see Theorem 12.1). Roughly speaking, µ β is constructed so thatits β/ε -codimensional Hausdorff measure is equivalent to the measure ν on Z . ThusTheorem 1.1 follows immediately from combining Theorems 11.3 and 12.1.The smoothness exponent θ = 1 − β/εp > d -sets in unweighted R n , considered by Jonsson–Wallin [29], [30].We also show that for ν -a.e. z ∈ Z the trace Tr u ( z ) is achieved in three differentways, namely, as averaged pointwise limits (11.4) and (11.5), by Lebesgue pointintegral averages (11.17) and as a pointwise restriction from the Newtonian space N ,p ( X ε ).Our study includes p = 1 and shows that the Besov space B θ , ( Z ) is the tracespace of the Newtonian space N , ( X ε , µ β ) when β = (1 − θ ) ε = (1 − θ ) log α . This isin contrast to the result of Gagliardo [19] that the trace space of the Sobolev space W , (Ω) is L ( ∂ Ω) when Ω is a Lipschitz domain in R n . This trace operator isnonlinear, which is necessary according to a result due to Peetre [44]. See Mal´y [39,Section 7] and Mal´y–Shanmugalingam–Snipes [40, Theorem 1.2] for metric spaceanalogs of this. The key difference between the setting of [40] and the current paperis that in [40] the measure on ∂ Ω has codimension 1 relative to the measure on Ω,while we have codimension β/ε < p which precludes us from having β/ε = 1 when p = 1. See also [39, Section 7] for the importance of this difference.To prove the above theorem, for each choice of α, τ > < ε ≤ log α , weconstruct a hyperbolic filling X of the metric space Z . Roughly speaking, is aninfinite graph whose vertices correspond to maximal α − n -separated subsets of Z foreach positive integer n . The role of τ is to define nearness between two points ineach of these sets as in (3.2). We then equip X with the uniformized metric d ε ( x, y ) = inf γ Z γ e − εd ( · ,v ) ds, where d ( · , v ) denotes the graph distance to the root v of the hyperbolic fillingand the infimum is taken over all curves in X joining x to y .Along the way, we explore how the choice of parameters affects the structure ofthe hyperbolic fillings X of Z and their uniformizations X ε : • Hyperbolic fillings are Gromov hyperbolic for all α, τ > τ = 1 (Example 8.8). • The uniformization X ε is a uniform space for all ε ≤ log α (Theorem 5.1). • The boundary of the uniformization X ε , with ε = log α , is biLipschitz equiv-alent to Z when Z is compact (Proposition 4.4).Subsequently, for a doubling measure ν on Z , we construct a lift of ν to ameasure µ on X which is uniformly locally doubling and supports a uniformlylocal 1-Poincar´e inequality. We then show that for every β >
0, the correspondingweighted measure dµ β ( x ) = e − βd ( x,x ) dµ ( x ) ≃ dist ε ( x, ∂ ε X ) β/ε dµ ( x ) (1.1)is globally doubling and supports a global 1-Poincar´e inequality on X ε and itsclosure X ε , see Theorem 10.3. This gives us the flexibility to choose β for each0 < θ < p ≥ θ = 1 − β/pε and thus see the nonlocal Besov space B θp,p ( Z ) as the trace space of the Newtonian space N ,p ( X ε , µ β ), with the advantagethat the Newtonian energy is local, and that the theory for Newtonian spaces ismore developed than the theory for Besov spaces on metric spaces.Invoking the regularity properties of Newtonian spaces, we then easily obtainseveral regularity results for Besov functions on Z : Anders Bj¨orn, Jana Bj¨orn and Nageswari Shanmugalingam
Corollary 1.2.
Let Z be a compact metric space equipped with a doubling mea-sure ν . Then for every < θ < , Lipschitz functions are dense in B θp,p ( Z ) andevery function in B θp,p ( Z ) has a representative that is quasicontinuous with respectto the Besov capacity.If p > s ν /θ , where s ν is the growth exponent of ν from (10.13) , then functionsin B θp,p ( Z ) have H¨older continuous representatives.If ν is in addition reverse-doubling, then functions in B θp,p ( Z ) belong to L q ( Z ) for q = s ν p/ ( s ν − pθ ) and have L q -Lebesgue points outside a set of zero B θp,p ( Z ) -capacity. Our results apply also to compact subsets of R n . On R n , the correspondingSobolev-type higher integrability result is due to Peetre [43, Th´eor`eme 8.1], whileNetrusov [42, Proposition 1.4] obtained the Lebesgue point result for q < np/ ( n − pθ ). Even though R n is not compact, the above corollary allows us to improveupon Netrusov’s result in the Euclidean setting by including q = np/ ( n − pθ ), seeProposition 13.6. These results show that our exponent q is optimal.The above density and quasicontinuity results are known to hold when the Besovspace is defined in terms of atomic decompositions or sequences of fractional Haj laszgradients (see for example Han–M¨uller–Yang [23, Definition 5.29], Koskela–Yang–Zhou [34, Definitions 1.2 and 4.4] and Heikkinen–Koskela–Tuominen [24, Defini-tion 2.5, Theorems 1.1 and 1.2]). Such spaces coincide with our Definition 9.7 when Z is unbounded and “reverse-doubling”, by [34, Theorem 4.1]. In particular, thedefinitions are equivalent if Z is uniformly perfect with ν doubling, for examplewhen Z = R n . With our assumptions on Z , it is not clear whether those definitionsagree with the nonlocal integral definition considered here. Note that the integral inDefinition 9.7 coincides with the ones defining the classical fractional Sobolev spacesin Euclidean spaces and is naturally related to nonlocal minimization problems forthe fractional p -Laplacian, as in Caffarelli–Silvestre [17] and Ferrari–Franchi [18].Higher integrability and H¨older continuity of Besov functions for large p ap-pears also in Mal´y [39, Corollary 3.18], where it is obtained as a consequence ofembeddings into Haj lasz–Sobolev spaces, provided by Lemma 6.1 in Gogatishvili–Koskela–Shanmugalingam [21]. Our approach based on hyperbolic fillings is differ-ent. Traces of Newtonian functions on uniform domains in metric spaces are alsostudied in [39, Theorem 1.1] by means of Lebesgue point averages and Poincar´einequalities. Our proof in the setting of hyperbolic fillings is more direct and ratherelementary (albeit a bit lengthy) and is based only on the basic properties of uppergradients. In particular, the Poincar´e inequality is not used. Moreover, we showthat for any compact doubling space Z , the Besov space B θp,p ( Z ) with 0 < θ < N ,p ( X ε ), see Theorem 1.1.Let us compare our definition of Besov spaces with some other function spaces onboundaries of hyperbolic fillings. Assuming Z to be uniformly perfect and Ahlfors Q -regular, the Besov space considered in Bourdon–Pajot [13] corresponds to our B θp,p ( Z ) with θ = Q/p and is shown to be isomorphically equivalent to the firstcohomology group ℓ p H ( X ). For Ahlfors Q -regular spaces Z , the papers [10], [11]and [48] by Bonk, Saksman and Soto define certain function spaces on Z by meansof Poisson-type extensions as in our Theorem 12.1, and using the counting mea-sure on the collection of all edges in the hyperbolic filling. In [10] they show thatif Z supports a Q -Poincar´e inequality then their space A p ( Z ) coincides with theHaj lasz–Sobolev space M ,Q ( Z ). The function spaces considered in [11] are ofTriebel–Lizorkin type, while the ones in [48] are identified with the Haj lasz–Besovspaces ˙ N sp,q ( Z ), defined by atomic decompositions in Koskela–Yang–Zhou [34, Def-inition 1.2].While these results are interesting, from our point of view it is somewhat unsat- xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 5 isfactory that the energy of functions considered in [10], [11] and [48] does not takeinto full account the measure ν on Z and that ν is not related to a measure on thehyperbolic filling.Unlike in [48], our definition of Besov spaces is based on integrals directly onthe metric space Z , rather than on sequence spaces, see Definition 9.7. Moreover,the smoothness of the corresponding Poisson extension on the hyperbolic filling X ,is controlled by a measure on X that is compatible with the measure ν on Z .The structure of this paper is as follows. The necessary background related tometric notions and Gromov hyperbolic spaces is given in Section 2, while notionsrelated to Newtonian and Besov spaces are given in Section 9.In Section 3 we describe the construction of the hyperbolic filling X of a generalbounded metric space Z , associated with the parameters α, τ >
1, and show thatit indeed forms a Gromov hyperbolic space. Subsequently, in Section 4 we describethe uniformization X ε of X , with parameter ε >
0, in the style of Bonk–Heinonen–Koskela [8], adapted to the setting of hyperbolic fillings. In this section we alsoexplore links between the boundary ∂ ε X of the uniformized space and the originalmetric space Z . In particular, the results in Section 4 show why the bound ε ≤ log α is natural.The primary goal of Section 5 is to prove that the uniformization X ε of X yieldsa uniform space when ε ≤ log α . The general results of [8] imply that X ε is auniform space for sufficiently small ε , but our direct proof for hyperbolic fillingscovers all ε ≤ log α , which is vital for our further results. Observe that generalGromov hyperbolic spaces do not always yield a uniform space when uniformized,see for example Lindquist–Shanmugalingam [35, Section 4].Given that all α, τ > Z , it is worth exploringhow the choice of these parameters affects the structure of the hyperbolic filling.The rough similarity between an arbitrary locally compact roughly starlike Gromovhyperbolic space X and the hyperbolic filling b X of its uniformized boundary ∂ ε X isfor small ε proved in Section 6, without any limitations on α in terms of ε . Trees andhyperbolic fillings of their uniformized boundaries, as well as some counterexamples,are considered in Section 7. In Section 8 we show that if τ ≥ ( α + 1) / ( α − Z equipped with adoubling measure ν , and its hyperbolic filling X as well as the uniformization X ε for 0 < ε ≤ log α . Following the description of notions related to Newtonian andBesov spaces given in Section 9, we describe in Section 10 our method of lifting upthe measure ν on Z to a measure µ on X . In that section we also show that theuniformization µ β of the measure µ , given for β > X ε and itscompletion X ε .The trace and extension results from Theorem 1.1 are proved in their specificforms as Theorems 11.1, 11.3 and 12.1, respectively. Finally, the results stated inCorollary 1.2 are obtained in Section 13 by exploiting the perspective of the Besovspaces as traces of Newtonian spaces.The third author communicated the results of this paper with Butler, who madeuse of some of the techniques of this paper together with the tools of the Busemanfunction to independently derive some of the results we obtain in Sections 3–5,with a focus on unbounded doubling metric spaces, see [15]. We do not addressthe issue of unbounded doubling metric spaces here, but the interested readers mayconsult [15]. However, his construction of the hyperbolic filling differs slightly from Anders Bj¨orn, Jana Bj¨orn and Nageswari Shanmugalingam ours in that he requires (3.4) instead of (3.3). The results in [15] also require that τ ≥ min (cid:26) , − /α (cid:27) (the parameter a in [15] corresponds to our 1 /α ), but we do not require any suchconstraint except in Section 8. Acknowledgement.
Parts of this research project were conducted during 2017and 2018 when N. S. was a guest professor at Link¨oping University, partially fundedby the Knut and Alice Wallenberg Foundation, and during the parts of 2019 whenA. B. and J. B. were Taft Scholars at the University of Cincinnati. The authorswould like to thank these institutions for their kind support and hospitality. A. B.and J. B. were partially supported by the Swedish Research Council grants 2016-03424 resp. 621-2014-3974 and 2018-04106. N. S. was partially supported by theNational Science Foundation (U.S.A.) grants DMS-1500440 and DMS-1800161.
2. Gromov hyperbolic spaces
In this section we will introduce Gromov hyperbolic spaces and uniform spaces anddiscuss relevant background results. In the later part of the paper we will needbackground results on upper gradients, Newtonian (Sobolev) spaces, Besov spaces,Poincar´e inequalities etc. This background discussion will be provided in Section 9.A curve is a continuous mapping from an interval. Unless stated otherwise,we will only consider curves which are defined on compact intervals. We denotethe length of a curve γ by ℓ ( γ ), and a curve is rectifiable if it has finite length.Rectifiable curves can be parametrized by arc length ds .A metric space X = ( X, d ) is geodesic if for each x, y ∈ X there is a curve γ with end points x and y and length ℓ ( γ ) = d ( x, y ). X is a length space if d ( x, y ) = inf γ ℓ ( γ ) for all x, y ∈ X, where the infimum is taken over all curves γ from x to y .A metric space is proper if all closed bounded sets are compact. We denoteballs in X by B ( x, r ) = { y ∈ X : d ( y, x ) < r } and the scaled concentric ball by λB ( x, r ) = B ( x, λr ). In metric spaces it can happen that balls with different centersand/or radii denote the same set. We will however adopt the convention that a ballcomes with a predetermined center and radius. Sometimes (especially when dealingwith several different spaces simultaneously) we will write B X and d X to indicatethat these notions are taken with respect to the metric space X . When we say that x ∈ γ we mean that x = γ ( t ) for some t . If γ is noninjective, then this t may notbe unique, but we are always implicitly referring to a specific such t . If x , x ∈ γ ,then γ x ,x denotes the subcurve between x and x . Definition 2.1.
A complete unbounded geodesic metric space X is Gromov hyper-bolic if there is a hyperbolicity constant δ ≥ x, y ], [ y, z ] and[ z, x ] are geodesics in X , every point w ∈ [ x, y ] lies within a distance δ of [ y, z ] ∪ [ z, x ].The ideal Gromov hyperbolic space is a metric tree, which is Gromov hyperbolicwith δ = 0. A metric tree is a tree where each edge is considered to be a geodesicof unit length. Definition 2.2.
An unbounded metric space X is roughly starlike if there are some x ∈ X and M > x ∈ X there is a geodesic ray γ in X , startingfrom x , such that dist( x, γ ) ≤ M . A geodesic ray is a curve γ : [0 , ∞ ) → X withinfinite length such that γ | [0 ,t ] is a geodesic for each t > xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 7 If X is a roughly starlike Gromov hyperbolic space, then the roughly starlikecondition holds for every choice of x , although M may change. Definition 2.3.
A nonempty open set Ω X in a metric space X is an A -uniformdomain , with A ≥
1, if for every pair x, y ∈ Ω there is a rectifiable arc lengthparametrized curve γ : [0 , ℓ ( γ )] → Ω with γ (0) = x and γ ( ℓ ( γ )) = y such that ℓ ( γ ) ≤ Ad ( x, y ) and d Ω ( γ ( t )) ≥ A min { t, ℓ ( γ ) − t } for 0 ≤ t ≤ ℓ ( γ ) , (2.1)where d Ω ( z ) = dist( z, X \ Ω) , z ∈ Ω . The curve γ is said to be an A -uniform curve . A noncomplete metric space (Ω , d )is A -uniform if it is an A -uniform domain in its completion.The completion of a locally compact uniform space is always proper, by Propo-sition 2.20 in Bonk–Heinonen–Koskela [8]. Unlike the definition used in [8], we donot require uniform spaces to be locally compact. We assume for the rest of this section that X is a roughly starlike Gromov δ -hyperbolic space. We also fix a point x ∈ X and let M be the constant in theroughly starlike condition with respect to x . The point x will serve as a center for the uniformization X ε of X . FollowingBonk–Heinonen–Koskela [8], we define, for a fixed ε >
0, the uniformized metric d ε on X as d ε ( x, y ) = inf γ Z γ ρ ε ds, where ρ ε ( x ) = e − εd ( x,x ) and the infimum is taken over all rectifiable curves γ in X joining x to y . Note thatif γ is a compact curve in X , then ρ ε is bounded from above and away from 0 on γ , and in particular γ is rectifiable with respect to d ε if and only if it is rectifiablewith respect to d . Definition 2.4.
The set X , equipped with the metric d ε , is denoted by X ε andcalled the uniformization of X , even when we do not know whether it is a uniformspace. We let X ε be the completion of X ε , and let ∂ ε X = X ε \ X ε be the uniformizedboundary of X ε (or X ).The uniformization X ε need not be a uniform space, as shown in Lindquist–Shanmugalingam [35, Section 4]. If X is locally compact and ε is sufficiently small,then ∂ ε X as a set is independent of ε and depends only on the Gromov hyperbolicstructure of X , see e.g. [8, Section 3]. The notation adopted in [8] is ∂ G X . Onthe other hand, if ε is large, then it is possible for ∂ ε X to change, see for exampleProposition 4.1 below.When writing e.g. B ε , diam ε and dist ε , the subscript ε indicates that thesenotions are taken with respect to ( X ε , d ε ). We also define d ε ( x ) = dist ε ( x, ∂ ε X ) . The length of the curve γ with respect to d ε is denoted by ℓ ε ( γ ). The arc length ds ε with respect to d ε satisfies ds ε = ρ ε ds. It follows that X ε is a length space, and thus also X ε is a length space. By a directcalculation (or [8, (4.3)]), diam ε X ε = diam ε X ε ≤ /ε .The following important theorem is due to Bonk–Heinonen–Koskela [8]; see [5,Theorem 2.6] for this version. By the Hopf–Rinow theorem (see Gromov [22, p. 9]for a suitable version), X is proper if and only if X is locally compact. Anders Bj¨orn, Jana Bj¨orn and Nageswari Shanmugalingam
Theorem 2.5.
Assume that X is locally compact. There is a constant ε ( δ ) > ,determined by δ alone, such that if < ε ≤ ε ( δ ) , then X ε is an A -uniform spacefor some A depending only on δ . Moreover, X ε is a compact geodesic space.If δ = 0 , then ε (0) can be chosen arbitrarily large. There is also a converse, again due to Bonk–Heinonen–Koskela [8]. Namely, if(
Y, d ) is a locally compact uniform space, then equipping Y with the quasihyperbolicmetric k Y gives a Gromov hyperbolic space, where k Y ( x, y ) := inf γ Z γ ds ( t )dist d ( γ ( t ) , ∂Y ) for x, y ∈ Y, with the infimum taken over all rectifiable curves γ in Y with end points x and y .We recall, for further reference, the following key estimates from [8]. Lemma 2.6. ([8, Lemma 4.16])
Assume that X is locally compact. Let ε > . If x ∈ X , then e − εd ( x,x ) eε ≤ dist ε ( x, ∂ ε X ) =: d ε ( x ) ≤ C e − εd ( x,x ) ε , (2.2) where C = 2 e εM − . In particular, εd ε ( x ) ≃ ρ ε ( x ) , and x → ∂ ε X with respect to d ε if and only if d ( x, x ) → ∞ . Corollary 2.7. ([5, Corollary 2.9 and its proof])
Assume that X is locally compactand that < ε ≤ ε ( δ ) , where ε ( δ ) is given by Theorem . Let x, y ∈ X . Then d ε ( x, y ) d ε ( x ) d ε ( y ) . exp( εd ( x, y )) . If εd ( x, y ) ≥ then exp( εd ( x, y )) ≃ d ε ( x, y ) d ε ( x ) d ε ( y ) , where the comparison constants depend only on δ , M and ε ( δ ) . Here and later, we write a . b if there is an implicit constant C > a ≤ Cb , and analogously a & b if b . a . We also use the notation a ≃ b to mean a . b . a .In the later part of the paper we will equip uniformizations of Gromov hyperbolicspaces and their boundaries with doubling measures. In the first part of the paper,the following metric doubling condition will instead play a role in a few places, butfor most results no doubling assumption is needed.A metric space ( Y, d ) is doubling (or metrically doubling ) if there is a constant N d ≥ z ∈ Y and r >
0, the ball B ( z, r ) can be covered byat most N d number of balls with radius r . Doubling is a uniform version of totalboundedness. In particular, if Y is complete and doubling, then Y is also proper .A Borel regular measure µ on Y is doubling if there is a constant C > < µ (2 B ) ≤ Cµ ( B ) < ∞ for all balls B ⊂ Y . If Y carries a doubling measure, then Y is necessarily doubling. The converse isnot true in general. However, if Y is a complete doubling measure space, thenLuukkainen–Saksman [37] has shown that Y carries a doubling measure. For moreon doubling spaces and doubling measures, see Heinonen [25, Section 10.13]. xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 9
3. Construction of hyperbolic fillings
The technique of hyperbolic fillings of doubling metric spaces was first consideredin Buyalo–Schroeder [16, Chapter 6], and then used in Bourdon–Pajot [13] andBonk–Saksman [10]. The constructions are different in these papers, see below.We construct the hyperbolic filling as follows: We assume that a bounded metricspace Z is given, and fix the parameters α, τ > z ∈ Z . By scalingwe can assume that 0 ≤ diam Z <
1. As mentioned in Section 2, we later want toequip Z with a doubling measure, but to begin with no such requirement is needed.We set A = { z } and note that Z = B Z ( z , n we can choose a maximal α − n -separated set A n ⊂ Z such that A n ⊂ A m when m ≥ n ≥
0. A set A ⊂ Z is α − n -separated if d Z ( z, z ′ ) ≥ α − n whenever z, z ′ ∈ A are distinct. Then the balls B Z ( z, α − n ), z ∈ A n , are pairwise disjoint. Since A n is maximal, the balls B Z ( z, α − n ), z ∈ A n , cover Z . Here and from now on, n and m will always be nonnegative integers.Next, we define the “vertex set” V = ∞ [ n =0 V n , where V n = { ( x, n ) : x ∈ A n } . (3.1)Note that a point x ∈ A n belongs to A k for all k ≥ n , and so shows up asthe first coordinate in infinitely many points in V . Given two different vertices( x, n ) , ( y, m ) ∈ V , we say that ( x, n ) is a neighbor of ( y, m ) (denoted ( x, n ) ∼ ( y, m ))if and only if | n − m | ≤ τ B Z ( x, α − n ) ∩ τ B Z ( y, α − m ) = ∅ , if m = n, (3.2) B Z ( x, α − n ) ∩ B Z ( y, α − m ) = ∅ , if m = n ± . (3.3)We let the hyperbolic filling X be the graph formed by the vertex set V togetherwith the above neighbor relation (edges), and consider X to be a metric graph wherethe edges are unit intervals. As usual for graphs, we do not consider a vertex tobe its own neighbor. The distance between two points in X is the length of theshortest curve between them. Since X is a metric graph, it is easy to see that theseshortest curves exist, and thus X is a geodesic space.If ( x, n ) ∼ ( y, n + 1) we say that ( y, n + 1) is a child of ( x, n ) while ( x, n ) is a parent of ( y, n + 1). (We use this terminology also for rooted trees.) In general,each vertex has at least one child, and all vertices but for the root v = ( z ,
0) haveat least one parent. An edge ( x, n ) ∼ ( y, m ) is horizontal if m = n and vertical if m = n ± X is always a Gromov hyperbolic space,but first we compare our construction with those in Buyalo–Schroeder [16, Chap-ter 6], Bourdon–Pajot [13] and Bonk–Saksman [10]. In [13], Bourdon and Pajot usethe same construction as we do with α = e and τ = 1. It is pointed out in [10] thatthe choice τ = 1 causes problems in the proof of the hyperbolicity of the “hyperbolicfilling”, more specifically in the proof of [13, Lemme 2.2]. Indeed, in Example 8.8we construct a “hyperbolic filling” with τ = 1 and α = 2 which is not a Gromovhyperbolic space.According to Bonk–Saksman [10], it is enough to enlarge the balls with a factor >
1, but they make the specific choices α = τ = 2. Their construction is howeverslightly different from ours: Instead of (3.3), they require that (with τ = 2) τ B Z ( x, α − n ) ∩ τ B Z ( y, α − m ) = ∅ , even if m = n ± . (3.4)Thus, the hyperbolic fillings in [10] contain more vertical edges than those consideredin this paper (with α = τ = 2). Buyalo and Schroeder, in [16, Chapter 6], use a similar construction with α ≥ r = 1 /α ≤ in their notation) and τ = 2, but impose a different conditionwhen m = n ±
1, namely (with τ = 2) τ B Z ( y, α − m ) ⊂ τ B Z ( x, α − n ) , if m = n + 1 . Buyalo and Schroeder show that their hyperbolic filling is Gromov hyperbolic. Bonkand Saksman [10] refer to Bourdon–Pajot [13] for a proof, but mention that theproof in [13] is problematic for τ = 1, as considered in [13]. Both [13] and [10] havestronger assumptions on Z than here.When τ ≥ ( α + 1) / ( α − τ > Z is a Cantor set, obtained as theuniformized boundary of an infinite tree and equipped with the induced ultrametric,then its hyperbolic filling with the choice of τ < α gives back the original tree,whereas the choice of τ ≥ α does not give a tree as the hyperbolic filling of Z ,see Section 7. On the other hand, the estimates regarding traces and extensionsof Sobolev and Besov spaces in Sections 11–12 (related to the uniformization ofhyperbolic fillings) are not affected by the precise values of α and τ . In the rest of the section, we assume that Z is a metric space with diam Z < and that X is a hyperbolic filling, as constructed above, with parameters α, τ > . We consider the projection maps π : V → Z and π : V → { , , , ... } givenby π (( x, n )) = x and π (( x, n )) = n . We also set v := ( z ,
0) and use the
Gromovproduct ( v | w ) v = [ d X ( v , v ) + d X ( v , w ) − d X ( v, w )] , v, w ∈ V. It follows easily from the construction that the hyperbolic filling is connected.The following lemma is a more precise version of this.
Lemma 3.1.
For all v ∈ V we have d X ( v, v ) = π ( v ) . In particular, V is connectedin the graph sense, and X in the metric sense.Proof. The first claim is clear if v = v . So suppose that v = ( x, n ) for some positiveinteger n . By the construction of A j , there are x j ∈ A j , such that x ∈ B Z ( x j , α − j ), j = 0 , , ... , n . In particular, x = x n , v j := ( x j , j ) ∈ V and x ∈ B Z ( x j − , α − ( j − ) ∩ B Z ( x j , α − j )for j = 1 , ... , n . It follows that v ∼ v ∼ ... ∼ v n , and thus d X ( v, v ) ≤ n . As allother paths connecting these two points have length at least n , we have the requiredconclusion of the first claim. The last part follows directly.Using that every vertex has at least one child, the following consequence of theconstruction in the proof of Lemma 3.1 is immediate. We will use similar propertiesmany times in this paper without further ado. Corollary 3.2. (a)
For every vertex v there is a geodesic ray starting at v andcontaining v . (b) Every geodesic ray starting at v consists solely of vertical edges. (c) Any geodesic from any x ∈ X to the root v contains at most a half of ahorizontal edge. (d) X is roughly starlike with M = . Next, we provide a proof of the hyperbolicity for all parameters α, τ >
1. Theideas are similar to those in [13] and [16]. In particular the following lemma wasinspired by [13, Lemme 2.2]. As mentioned above, when τ = 1 it is possible for the“hyperbolic filling” to be nonhyperbolic, see Example 8.8 below. xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 11 Lemma 3.3.
Let v = ( z, n ) and w = ( y, m ) be two vertices in X . Then α − ( v | w ) v ≃ d Z ( z, y ) + α − n + α − m , with comparison constants depending only on α and τ .Proof. Without loss of generality, we can assume that n ≤ m . If z = y then d X ( v, w ) = m − n and therefore ( v | w ) v = ( n + m − ( m − n )) = n , and so thestatement holds in this case. Assume therefore that z = y .Let l be the smallest nonnegative integer such that α − l ≤ τ −
1, and k bethe smallest nonnegative integer such that α − k − < d Z ( z, y ). For j = 0 , , ... , let z j , y j ∈ A j be such that d Z ( z, z j ) < α − j and d Z ( y, y j ) < α − j . Clearly, we can choose z j = z for j ≥ n and y j = y for j ≥ m . We shall distinguishtwo cases:If k := min { k − l, n } ≥
0, then α k − k ≤ α − l ≤ τ −
1, and the triangle inequalityshows that d Z ( z, y k ) ≤ d Z ( z, y ) + d Z ( y, y k ) < α − k + α − k = ( α k − k + 1) α − k ≤ τ α − k . Hence z ∈ τ B Z ( z k , α − k ) ∩ τ B Z ( y k , α − k ), from which it follows that( z, n ) ∼ ( z n − , n − ∼ ... ∼ ( z k , k ) ∼ ( y k , k ) ∼ ... ∼ ( y n − , m − ∼ ( y, m )where the middle edge may collapse into a single vertex. Thus, d X ( v, w ) ≤ n + m + 1 − k , and consequently, ( v | w ) v = ( n + m − d X ( v, w )) ≥ k − . If k < v | w ) v ≥ > k . In both cases we thus have that α − ( v | w ) v ≤ α − k +1 / ≤ α / ( α l − k + α − n ) ≤ α l +3 / ( d Z ( z, y ) + α − n + α − m ) . Note that l only depends on α and τ .Conversely, let w ∼ w ∼ ... ∼ w N be a geodesic from v = w to w = w N . Notethat d X ( v, w ) = N ≥ m − n and ( v | w ) v = ( n + m − N ) . Moreover, by the construction of the hyperbolic filling, π ( w j ) ≥ n − j and π ( w N − i ) ≥ m − i for all i, j = 0 , , ... , N . Therefore α − π ( w j ) ≤ α j − n and α − π ( w N − i ) ≤ α i − m . The triangle inequality then yields that for all k = 0 , , ... , N , d Z ( z, y ) + α − n + α − m ≤ N X j =1 d Z ( π ( w j − ) , π ( w j )) + α − n + α − m < α − n + N X j =1 ( τ α − π ( w j − ) + τ α − π ( w j ) ) + α − m ≤ τ k − X j =0 α j − n + 2 τ N − k X i =0 α i − m < τα − α k − n + α N − k − m +1 ) , where the sum P k − j =0 is empty when k = 0. Choosing k to be the smallest integer ≥ ( N + n − m ) gives that d Z ( z, y ) + α − n + α − m < τ αα − α ( N − n − m ) / = 4 τ αα − α − ( v | w ) v . Theorem 3.4.
There is a constant C ≥ , depending only on α and τ , such that if u , v and w are three vertices in V , then ( v | w ) v ≥ min { ( v | u ) v , ( w | u ) v } − C. (3.5) In particular, X is Gromov hyperbolic.Proof. It suffices to prove (3.5) since Gromov hyperbolicity is equivalent to it, seeBonk–Heinonen–Koskela [8] or Bridson–Haefliger [14, p. 411, Proposition 1.22].Let v = ( z, n ), w = ( y, m ) and u = ( x, k ). Then clearly, d Z ( z, y ) + α − n + α − m ≤ ( d Z ( z, x ) + α − n + α − k ) + ( d Z ( x, y ) + α − k + α − m ) , and Lemma 3.3 implies that α − ( v | w ) v . α − ( v | u ) v + α − ( u | w ) v ≤ α − min { ( v | u ) v , ( u | w ) v } . Taking logarithms concludes the proof.
4. The uniformized boundary ∂ ε X In this section, we assume that Z is a metric space with diam Z < , and let X bea hyperbolic filling of Z with parameters α, τ > . In this section we will look at how Z relates to the uniformized boundary ∂ ε X of X , see Section 2 for the definitions. We use the root v = ( z ,
0) as the uni-formization center x . We will show that if ε ≤ log α and Z is complete, then ∂ ε X is snow-flake equivalent to Z , with exponent σ = ε/ log α , see Proposition 4.4 forfurther details. In particular, ∂ ε X and Z are biLipschitz equivalent if ε = log α .Before showing the equivalence of ∂ ε X and Z , we take a look at the case ε > log α . Towards the end of the section we will also study how the degree of thevertices in X depends on properties of Z . In this section we are not concerned withwhether the uniformization X ε is a uniform domain or not. This question will beconsidered in Section 5.Note that ρ ε ≃ ρ ε ( v ) on every edge [ v, w ] ⊂ X and thus, using also the basicfacts from Corollary 3.2, for all x ∈ X , d ε ( x ) ≃ Z ∞ d X ( x,v ) ρ ε ds = 1 ε ρ ε ( x ) , (4.1)with equality if x is a vertex.The following result shows that if ε > log α , then ∂ ε X often becomes just onepoint. Proposition 4.1.
Assume that there is
L < ∞ such that any two points in the d Z -completion Z of Z can be connected by a curve in Z of length at most L . If ε > log α , then ∂ ε X consists of just one point. This shows that when ε > log α and Z has at least two points, there is nonatural connection between Z and ∂ ε X . On the other hand, it is easy to see thatif Z consists of finitely many points, then ∂ ε X is also finite and there is a naturalbiLipschitz map between these sets. (We leave the details regarding finite sets Z tothe interested reader.) In particular, some connectivity assumption is necessary inProposition 4.1. xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 13 Proof.
Let F n = { x ∈ X : d X ( x, v ) ≤ n } and let x, x ′ ∈ X \ F n be arbitrary.Then there are d X -geodesics from x and x ′ to v which contain vertices v, v ′ ∈ V n ,respectively. By the connectivity assumption on Z , there is a sequence { w j } mj =0 ofpoints in Z such that v = ( w , n ), v ′ = ( w m , n ), m ≤ Lα n / ( τ −
1) and (if m ≥ d Z ( w j , w j − ) ≤ ( τ − α − n , j = 1 , ... , m. For each j = 0 , ... , m there are points w ′ j ∈ Z and z j ∈ A n such that d Z ( w ′ j , w j ) ≤ ( τ − α − n and d Z ( w ′ j , z j ) < α − n . We can choose z = w ′ = w and z m = w ′ m = w m . As d Z ( w ′ j , z j − ) ≤ d Z ( w ′ j , w j ) + d Z ( w j , w j − ) + d Z ( w j − , w ′ j − ) + d Z ( w ′ j − , z j − ) < ( τ − α − n + α − n = τ α − n when j ≥
1, we see that v = ( z , n ) ∼ ( z , n ) ∼ ... ∼ ( z m , n ) = v ′ . It follows that d ε ( x, x ′ ) ≤ d ε ( x, v ) + d ε ( v, v ′ ) + d ε ( v ′ , x ′ ) < Z ∞ n e − εt dt + me − εn < e − εn ε + e − εn Lα n τ − e − εn ε + Lτ − e n (log α − ε ) . Taking supremum over all x, x ′ ∈ X \ F n shows thatdiam ε ∂ ε X ≤ diam ε ( X \ F n ) → , as n → ∞ , since ε > log α .The easiest example of a space Z applicable in Proposition 4.1 is Z = [0 , ]. Thefollowing example shows that it is possible to apply this proposition to a noncompactcomplete space. Example 4.2.
Let Z consist of countably many copies of [0 , ], all glued togetherat 0, and equipped with the inner length metric, so that Z is geodesic. It is easy tosee that Z is noncompact and complete, and that diam Z = . Thus Proposition 4.1applies.Before proceeding, we deduce the following lemma which will be used severaltimes in this paper. Lemma 4.3.
Assume that z, y ∈ Z are such that d Z ( z, y ) ≤ α − k for some nonneg-ative integer k . For j = 0 , , ... , let z j , y j ∈ A j be such that d Z ( z, z j ) < α − j and d Z ( y, y j ) < α − j . Then v ∼ ( z , ∼ ... ∼ ( z k , k ) ∼ ( z k +1 , k + 1) ∼ ... . (4.2) Let l be the smallest nonnegative integer such that α − l ≤ τ − . Then, for any m, n ≥ h := max { k − l, } , ( z n , n ) ∼ ... ∼ ( z h , h ) ∼ ( y h , h ) ∼ ... ∼ ( y m , m ) , (4.3) where the middle edge collapses into a single vertex if z h = y h . This path γ haslengths ℓ X ( γ ) and ℓ ε ( γ ) ( with respect to d X resp. d ε ) satisfying d X (( z n , n ) , ( y m , m )) ≤ ℓ X ( γ ) ≤ n + m + 1 − h ≤ n + m + 1 + 2 l − k, (4.4) d ε (( z n , n ) , ( y m , m )) ≤ ℓ ε ( γ ) ≤ ε e − εh ≤ ε e − ε ( k − l ) . Note that z j and y j exist by the construction of A j , and that y = z with z being the unique element in A as before. Proof.
Since z ∈ B Z ( z j , α − j ) ∩ B Z ( z j +1 , α − ( j +1) ) for each j = 0 , , ... , we directlysee that (4.2) holds. By the triangle inequality and the choice of l we see that d Z ( z, y h ) ≤ d Z ( z, y ) + d Z ( y, y h ) < α − k + α l − k ≤ τ α l − k . Moreover, z ∈ τ B Z ( z h , α − h ) and so ( z h , h ) ∼ ( y h , h ) or z h = y h . Therefore (4.3)follows from (4.2) (and the corresponding path for y ). The estimate (4.4) followsdirectly from (4.3).To estimate ℓ ε ( γ ), recall that by Lemma 3.1, d X ( v, v ) = π ( v ) for all v ∈ V and that ρ ε ( x ) = e − εd X ( x,v ) for all x ∈ X . Hence ℓ ε ( γ ) ≤ Z nh e − εt dt + Z mh e − εt dt + 2 Z h +1 / h e − εt dt ≤ ε e − εh ≤ ε e − ε ( k − l ) , where the last integral estimates the d ε -length of the (possibly collapsed) horizontaledge ( z h , h ) ∼ ( y h , h ).We next show that ∂ ε X is snowflake equivalent to the completion Z of Z . Inparticular, ∂ ε X is biLipschitz equivalent to Z when ε = log α . Proposition 4.4.
Fix < ε ≤ log α . Then for all vertices v, w ∈ X , d Z ( π ( v ) , π ( w )) σ ≤ C d ε ( v, w ) , where C = (2 τ α ) σ and σ = ε log α ≤ . (4.5) Moreover, ∂ ε X is snowflake-equivalent to the completion Z of Z , that is, there is anatural homeomorphism Ψ : Z → ∂ ε X such that for all z, y ∈ Z , d Z ( z, y ) σ C ≤ d ε (Ψ( z ) , Ψ( y )) ≤ C d Z ( z, y ) σ , (4.6) where C = 4 α ( l +1) σ /ε and l is the smallest nonnegative integer such that α − l ≤ τ − .Proof. Let w ∼ w ∼ ... ∼ w N be a path γ in X connecting w = v to w N = w . We can assume without lossof generality that π ( w N ) ≤ π ( w ). Then by the construction of the hyperbolicfilling, d Z ( π ( w ) , π ( w N )) ≤ N X i =1 d Z ( π ( w i − ) , π ( w i )) ≤ τ N X i =1 α − π ( w i ) . (4.7)Moreover, for each i , ℓ ε ([ w i − , w i ]) ≥ Z π ( w i )+1 π ( w i ) e − εt dt > e − ε ( π ( w i )+1) = e − ε ( α − π ( w i ) ) σ . Summing over all i and using the elementary inequality ( P Ni =1 a i ) σ ≤ P Ni =1 a σi for σ ≤
1, together with (4.7), yields ℓ ε ( γ ) = N X i =1 ℓ ε ([ w i − , w i ]) ≥ e − ε N X i =1 ( α − π ( w i ) ) σ ≥ e − ε (cid:18) d Z ( π ( w ) , π ( w N ))2 τ (cid:19) σ . xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 15 Taking the infimum over all such paths γ gives (4.5).Let z ∈ Z and find a sequence z j ∈ A j such that d Z ( z j , z ) < α − j . Then fromLemma 4.3 we see that v ∼ ( z , ∼ ... ∼ ( z j , j ) ∼ ( z j +1 , j + 1) ∼ ... . Moreover,when i > j , d ε (( z i , i ) , ( z j , j )) ≤ Z ∞ j e − εt dt = e − εj ε . It follows that { ( z j , j ) } is a Cauchy sequence in X ε . Set Ψ( z ) = lim j →∞ ( z j , j ).From the construction of X ε it is clear that Ψ( z ) / ∈ X ε , and so this point belongs to ∂ ε X . If z ∗ j ∈ A j is such that d Z ( z ∗ j , z ) < α − j , then z ∈ B Z ( z j , α − j ) ∩ B Z ( z ∗ j , α − j )and thus ( z ∗ j , j ) ∼ ( z j , j ). Hence d ε (( z ∗ j , j ) , ( z j , j )) . e − εj and so lim j →∞ ( z ∗ j , j ) =lim j →∞ ( z j , j ), which shows that Ψ( z ) is well-defined and gives the map Ψ : Z → ∂ ε X .When z, y ∈ Z with z = y and k is a nonnegative integer such that α − k − The hyperbolic filling X has uniformly bounded degree if and onlyif Z is doubling.The uniformity and doubling constants depend only on α , τ and each other.Proof. Assume first that Z is doubling. Let ( x, n ) ∈ X and set A ( x, n ) ⊂ V to be thecollection of all neighbors of ( x, n ). For ( y, m ) ∈ A ( x, n ) we know that | m − n | ≤ τ B Z ( x, α − n ) ∩ τ B Z ( y, α − m ) is nonempty. Hence d Z ( x, y ) < τ ( α + 1) α − n andso y ∈ τ ( α + 1) B Z ( x, α − n ). Since Z is assumed to be doubling, there is a positiveinteger N independent of n , such that the ball τ ( α + 1) B Z ( x, α − n ) can be coveredby balls B , ... , B N of radius α − n − , see Heinonen [25, p. 81].Now, for each m ∈ { n − , n, n + 1 } , we have α − m ≥ α − n − and the balls B Z ( y, α − m ), y ∈ A m , are pairwise disjoint. It follows that each ball B j , j =1 , ... , N , can contain at most one point from A m . Hence, there are at most N such y ∈ A m satisfying ( y, m ) ∈ A ( x, n ). Since this is true for each m ∈ { n, n ± } , weconclude that the cardinality of A ( x, n ) is at most 3 N , that is, X is of uniformlybounded degree. Conversely, assume that X has a uniformly bounded degree. Let ζ ∈ Z and0 < r ≤ 1. Let k be the smallest nonnegative integer such that min { , r } ≤ α − k .For this choice of k , let n be the smallest integer such that n ≥ k and α − n ≤ r .Note that n ≤ k + l ′ for some l ′ depending only on α .Since the balls B Z ( z, α − n ) with z ∈ A n cover Z , for every ξ ∈ B Z ( ζ, r ) there is z ∈ A n such that ξ ∈ B Z ( z, α − n ) ⊂ B Z ( z, r ) . Moreover, d Z ( z, ζ ) ≤ d Z ( z, ξ ) + d Z ( ξ, ζ ) < α − n + r ≤ r. It follows that the balls B Z ( z, α − n ) with z ∈ A := A n ∩ B Z ( ζ, r ) cover B ( ζ, r ). Toestimate the cardinality of A , note that any two points z, z ′ ∈ A satisfy d Z ( z, z ′ ) ≤ min { , r } ≤ α − k . Lemma 4.3 and the above observation that n ≤ k + l ′ then implythat (with l as in Lemma 4.3), d X (( z, n ) , ( z ′ , n )) ≤ n + 1 + 2( l − k ) ≤ l + l ′ ) , which only depends on α and τ . By assumption there is a uniform bound on thedegrees in X and hence also on the number of vertices in balls with a fixed radius.Thus there is a uniform bound on the cardinality of A , i.e. Z is doubling. Proposition 4.6. Let ε > and let X ε be the uniformization of X , as defined inSection . Then the following are equivalent :(a) Z is totally bounded ;(b) each vertex layer V n ( as defined in (3.1)) is finite ;(c) every vertex in X has finite degree ;(d) X , and equivalently X ε , is locally compact ;(e) X ε is compact.If ε ≤ log α , then the following condition is also equivalent to those above :(f) ∂ ε X is compact.Moreover, X ε is geodesic whenever (a) – (e) hold. Note that we do not require X ε to be uniform. Since (a)–(c) are independentof ε , so are (d) and (e). Furthermore, (f) is also independent of ε provided that ε ≤ log α . Example 4.2 shows that (f) is not equivalent to the other statementswhen ε > log α . Proof. (a) ⇔ (b) It follows directly from the definition of total boundedness that Z is totally bounded if and only if all A n are finite sets, or equivalently all V n arefinite.(b) ⇒ (c) Let v = ( x, n ) be a vertex. Then all neighbors of v belong to the finiteset V n − ∪ V n ∪ V n +1 , i.e. v has finite degree. ¬ (b) ⇒ ¬ (c) Let m be the least index such that V m is infinite (which exists as(b) fails). As V = { v } , we must have m ≥ 1. Each vertex in V m has at leastone parent in V m − . As V m − is finite and V m is infinite, there must be a vertex in V m − which has infinitely many children, and hence has infinite degree.(c) ⇔ (d) This is easily seen to be true. Note that X and X ε have the sametopology, and are thus simultaneously locally compact or not.(b) ⇒ (e) Let F n = { x ∈ X : d X ( x, v ) ≤ n } . Since each V j is finite, it followsthat F n is a union of finitely many compact intervals and so is compact.Consider x ∈ X \ F n and let γ be a geodesic from x to v . As d X ( x, v ) > n ,there is some point v ∈ γ such that d X ( v, v ) = n . Since n is an integer, it followsthat v ∈ V n . Hence dist ε ( x, V n ) ≤ Z ∞ n e − εt dt = e − εn ε . xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 17 This inequality also holds for x ∈ ∂ ε X , since ∂ ε X ⊂ X \ F n (where the closure iswith respect to the d ε metric).Let η > n so that e − εn /ε < η . Then X ε \ F n ⊂ S y ∈ V n B ε ( y, η ),and as F n is compact and V n is finite we see that there is a finite η -net for X ε . Since η > X ε is totally bounded and thus compact (as it is complete bydefinition).(e) ⇒ (b) As V n ⊂ X ε , it is also compact with respect to the metric d ε . Themetrics d X and d ε are biLipschitz equivalent on V n and thus V n is compact alsowith respect to d X . Since distinct points in V n are at least a distance 1 apart, itfollows that V n is a finite set.Next, we assume that ε ≤ log α and consider (f).(e) ⇒ (f) This is trivial.(f) ⇒ (a) It follows from Proposition 4.4 that Z is homeomorphic to ∂ ε X , andis thus also compact. Hence Z is totally bounded.Finally, as X ε is a length space, it follows from Ascoli’s theorem that it isgeodesic if it is compact. 5. Uniformizing a hyperbolic filling with parame-ter ε ≤ log α In this section, we assume that Z is a metric space with diam Z < , and let X bea hyperbolic filling of Z with parameters α, τ > . The aim of this section is to show that the uniformization X ε is a uniformdomain when ε ≤ log α . (Recall Definition 2.3 of uniform spaces and uniformcurves.) Since X is Gromov hyperbolic (by Theorem 3.4), it follows from Bonk–Heinonen–Koskela [8, Theorem 2.6] (see Theorem 2.5) that X ε is a uniform space forsufficiently small ε > 0, when X is locally compact. In the later part of this paperwe are interested in uniformizing (a locally compact) X with respect to ε = log α ,and so we cannot rely on Theorem 2.5 or [8]. Therefore we provide a direct proofhere, which also avoids assuming local compactness. As we saw in Section 4, it isnatural to assume that ε ≤ log α in order for the boundary ∂ ε X to be homeomorphicto Z (or its completion Z if Z is not complete). Theorem 5.1. For all < ε ≤ log α , the uniformized space X ε is uniform with theuniformity constant depending only on α , τ and ε . Moreover, for all x ′ , x ′′ ∈ X , d ε ( x ′ , x ′′ ) ≃ e − ε ( x ′ | x ′′ ) v min { d X ( x ′ , x ′′ ) , } , (5.1) with comparison constants depending only on α , τ and ε .Proof. First we show that curves given by Lemma 4.3, connecting vertices in X , arequasiconvex curves (i.e. having length at most a constant multiple of the distancebetween their endpoints). These curves will be subsequently used to constructuniform curves connecting pairs of points that are “far apart”. Throughout theproof, we let l be the smallest nonnegative integer such that α − l ≤ τ − v = ( z, n ) and w = ( y, m ) be two distinct vertices in V with z ∈ A n ⊂ Z and y ∈ A m ⊂ Z . We can assume that n ≤ m . Let k be the largest nonnegativeinteger such that k ≤ n and d Z ( z, y ) ≤ α − k . Consider a curve γ connecting v to w ,as in (4.3) of Lemma 4.3. Then by Lemma 4.3 we have ℓ ε ( γ ) ≤ ε e − ε ( k − l ) = 4 e ε ( l +1) ε ( α − k − ) σ , where σ = ε log α ≤ . We now show that ℓ ε ( γ ) is comparable to d ε ( v, w ). If k < n then the comparisonfollows from the choice of k and from (4.5), which together imply that ℓ ε ( γ ) . ( α − k − ) σ < d Z ( z, y ) σ ≤ (2 τ α ) σ d ε ( v, w ) . On the other hand, if k = n , then any injective curve γ ′ connecting v to w startswith an edge v ∼ v ′ , and thus ℓ ε ( γ ′ ) ≥ d ε ( v, v ′ ) ≥ Z k +1 k e − εt dt ≥ e − ε ( k +1) = ( α − k − ) σ & ℓ ε ( γ ) . Taking infimum over all such curves γ ′ shows that even when k = n we have ℓ ε ( γ ) . d ε ( v, w ) . (5.2)Now assume that x ′ , x ′′ ∈ X are two arbitrary distinct points and consider aninjective curve b γ from x ′ to x ′′ with ℓ ε ( b γ ) < d ε ( x ′ , x ′′ ). If b γ contains at most onevertex, then d X ( x ′ , x ′′ ) < 2, and by (4.1), ℓ ε ( b γ ) ≃ d ε ( x ′ , x ′′ ) ≃ e − εd X ( x ′ ,x ) d X ( x ′ , x ′′ ) . d ε ( x ) for all x ∈ b γ, (5.3)and thus b γ is a uniform curve. By the triangle inequality,( x ′ | x ′′ ) v = [ d X ( x ′ , x ) + d X ( x ′′ , x ) − d X ( x ′ , x ′′ )] ≥ [ d X ( x ′ , x ) + d X ( x ′ , x ) − d X ( x ′ , x ′′ )] > d X ( x ′ , x ) − , and by the triangle inequality again, ( x ′ | x ′′ ) v ≤ d X ( x ′ , x ). Inserting this into (5.3)shows that (5.1) holds in this case.If b γ contains at least two vertices, then let v = ( z, n ) and w = ( y, m ) be thefirst resp. last vertex in b γ . Let k be the largest nonnegative integer such that both k ≤ min { n, m } and d Z ( z, y ) ≤ α − k . Let γ be a curve as described at the beginningof this proof, connecting the vertices v and w . The desired uniform curve e γ between x ′ and x ′′ is then obtained by appending the segments [ x ′ , v ] and [ w, x ′′ ] to γ . By(5.2), ℓ ε ( e γ ) ≤ d ε ( x ′ , v ) + ℓ ε ( γ ) + d ε ( w, x ′′ ) . d ε ( x ′ , v ) + d ε ( v, w ) + d ε ( w, x ′′ ) = ℓ ε ( b γ ) < d ε ( x ′ , x ′′ ) . We next show that e γ satisfies the twisted cone condition (2.1) in Definition 2.3.Note that e γ need not be injective and recall from Section 2 what y ∈ e γ and e γ y,y ′ mean in such a case.Let v ′ = ( z h , h ) ∈ γ and w ′ = ( y h , h ) ∈ γ , where h = max { k − l, } , z h and y h areas given for v and w by Lemma 4.3. Recall that γ consists of two vertical segments,one from v to v ′ and the other from w to w ′ , together with the (possibly collapsed)horizontal edge [ v ′ , w ′ ]. Let x ∈ e γ be arbitrary and consider the subcurve e γ x,x ′ of e γ from x to x ′ . We shall distinguish three basic situations and their symmetricequivalents. If x ∈ e γ x ′ ,v , then ℓ ε ( e γ x,x ′ ) = d ε ( x, x ′ ) ≤ e − ε ( n − ≃ d ε ( x ) . If x ∈ e γ v,v ′ , then e γ v,x is a vertical segment, and hence, using (4.1) and that π ( v ) = n ≥ d X ( x, v ), ℓ ε ( e γ x,x ′ ) = d ε ( x ′ , v ) + Z nd X ( x,v ) e − εt dt ≤ e − ε ( n − + Z ∞ d X ( x,v ) e − εt dt . d ε ( x ) . xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 19 If x ∈ e γ v ′ ,w ′ , then π ( v ′ ) = h ≤ n , and thus ℓ ε ( e γ x,x ′ ) ≤ d ε ( x ′ , v ) + Z nh e − εt dt + 2 Z h +1 / h e − εt dt ≤ e − ε ( n − + 3 Z ∞ h e − εt dt . e − εh ≃ d ε ( x ) . The case when x ∈ e γ w ′ ,x ′′ is treated similarly. Thus, e γ is a uniform curve.To prove (5.1) also in this case, note that by Lemma 4.3, and using that k is thelargest nonnegative integer ≤ min { n, m } such that d Z ( z, y ) ≤ α − k , d ε ( x ′ , x ′′ ) ≤ d ε ( x ′ , v ) + d ε ( v, w ) + d ε ( w, x ′′ ) . e − εn + e − εk + e − εm . d Z ( z, y ) σ + α − σn + α − σm . Conversely, (4.5) shows that2 d ε ( x ′ , x ′′ ) > ℓ ε ( b γ ) ≥ d ε ( v, w ) & d Z ( z, y ) σ . Since also d ε ( v, w ) & e − εn + e − εm = α − σn + α − σm , we get by Lemma 3.3 that d ε ( x ′ , x ′′ ) ≃ d Z ( z, y ) σ + α − σn + α − σm ≃ α − σ ( v | w ) v ≃ e − ε ( x ′ | x ′′ ) v . 6. Roughly similar equivalence In this section, we want to show that every locally compact roughly starlike Gro-mov hyperbolic space is roughly similar to any hyperbolic filling of its uniformizedboundary when the uniformization index ε is small enough to guarantee that theuniformized space is a uniform space.Let X be a locally compact roughly starlike Gromov hyperbolic space. For0 < ε ≤ ε ( δ ), where ε ( δ ) is given by Theorem 2.5, we uniformize X , with uni-formization point x , to obtain X ε equipped with the metric d ε and having boundary ∂ ε X . It will be convenient to use the scaled metricˆ d ε = εe d ε on X ε , and correspondingly d dist ε and [ diam ε . A consequence is that ˆ d ε ( x ) = 1 /e and 1 e ≤ [ diam ε X ≤ e < [ diam ε ∂ ε X ≤ /e < 1. Note that if X is the half-line, then ∂ ε X consistsof just one point and thus has diameter 0. Definition 6.1. Let ( W, d W ) and ( Y, d Y ) be metric spaces. A (not necessarilycontinuous) map Φ : W → Y is a rough similarity if there are C ≥ L ≥ Y is at most a distance C from Φ( W ), and for all x, x ′ ∈ W , Ld W ( x, x ′ ) − C ≤ d Y (Φ( x ) , Φ( x ′ )) ≤ Ld W ( x, x ′ ) + C. (6.2)We refer interested readers to Bonk–Schramm [12, Section 2] for more on roughsimilarity between Gromov hyperbolic spaces. Theorem 6.2. Assume that X is a locally compact roughly starlike Gromov hy-perbolic space and that < ε ≤ ε ( δ ) , where ε ( δ ) is given by Theorem . Let Z = ∂ ε X be the uniformized boundary of X equipped with the metric ˆ d ε . Let b X be any hyperbolic filling of Z , constructed with parameters α, τ > and the maximal α − n -separated sets A n ⊂ Z .Consider the mapping Φ : X → b X , defined for x ∈ X with α − n − < ˆ d ε ( x ) ≤ α − n , n = 0 , , ... , by Φ( x ) = ( z, n ) , where z is chosen to be a nearest point in A n to x , i.e. ˆ d ε ( x, z ) = d dist ε ( x, A n ) .Then Φ is a rough similarity with L = (log α ) /ε = 1 /σ and C depending onlyon α , τ , ε ( δ ) , δ and M . By combining Bonk–Schramm [12, Theorem 8.2] with Proposition 4.4 we knowthat X and b X are roughly similar if ε > ∂X ε issnowflake equivalent to the visual boundary of X as in Bonk–Heinonen–Koskela [8].However, as this rough similarity is obtained from [12] via comparison with the coneCon( ∂Y ε ), here we give a more direct construction of the rough similarity between X and b X . In so doing, we also demonstrate how the parameters α , τ and ε affectthe rough similarity constants.We chose to point out the dependence on ε ( δ ) separately in Theorem 6.2 eventhough ε ( δ ) is supposed to be determined solely by δ , because should a betterupper bound for ε in Theorem 2.5 and Corollary 2.7 be known in the future, theresult in this theorem will also be valid for the enlarged range of ε . Note that Z = ∂ ε X is compact by Theorem 2.5, and thus so is A n , which shows that thenearest points above exist. Proposition 7.3 shows that Z can be nondoubling.To prove this theorem, we need the following lemma. Let N be the smallestinteger ≥ / log α . For each n ≥ N we set S n = { x ∈ X ε : ˆ d ε ( x ) = α − n } . (6.3)By the choice of N we know that S n = ∅ . Lemma 6.3. Suppose that the assumptions in Theorem hold. Fixing n ≥ N ,let z ∈ ∂ ε X and let x be a nearest point in S n to z . Then α − n = ˆ d ε ( x ) ≤ ˆ d ε ( z, x ) ≤ α − n e. As noted above, X ε is compact. Therefore S n is also compact, and hence thenearest points referred to above do exist. Proof. Find a sequence z k ∈ X such that z k → z with respect to ˆ d ε . Since X isroughly starlike, there is a sequence of arc length parametrized d X -geodesic rays γ k : [0 , ∞ ) → X starting from x , and a sequence of points w k ∈ γ k , such thatdist X ( w k , z k ) ≤ M . For all y ∈ [ z k , w k ] (where [ z k , w k ] is any d X -geodesic from z k to w k ), d X ( y, x ) ≥ d X ( z k , x ) − M and hence ρ ε ( y ) ≤ e εM ρ ε ( z k ) . It follows thatˆ d ε ( z k , w k ) ≤ εe Z [ z k ,w k ] ρ ε ds ≤ M εe e εM ρ ε ( z k ) → , as k → ∞ , showing that w k → z . The ray γ k intersects S n and for sufficiently large k thereexists y k = γ k ( t k ) ∈ γ k ∩ S n such that w k ∈ γ k (( t k , ∞ )). As y k ∈ S n , using (2.2)we see that α − n = ˆ d ε ( y k ) = εe d ε ( y k ) ≥ e − εd X ( y k ,x ) e = e − εt k e . xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 21 Consequently, ˆ d ε ( w k , y k ) ≤ εe Z ∞ t k e − εt dt = e − εt k e ≤ α − n e. This finally implies, since w k → z , that d dist ε ( z, S n ) ≤ ˆ d ε ( z, y k ) ≤ ˆ d ε ( z, w k ) + ˆ d ε ( w k , y k ) ≤ ˆ d ε ( z, w k ) + α − n e → α − n e, as k → ∞ , which proves the last inequality in the statement of the lemma. The remaining(in)equalities are clear from the definitions. Proof of Theorem . Let x, x ′ ∈ X . Let n and m be the largest integers such thatˆ d ε ( x ) ≤ α − n and ˆ d ε ( x ′ ) ≤ α − m , respectively. Note that n, m ≥ v = Φ( x ) = ( z, n ) and w = Φ( x ′ ) = ( y, m ), with z ∈ A n and y ∈ A m . Note thatˆ d ε ( x, z ) < α − n and ˆ d ε ( x ′ , y ) < α − m . (6.4)The triangle inequality and Lemma 3.3 imply thatˆ d ε ( x, x ′ ) ≤ ˆ d ε ( z, y ) + 2( α − n + α − m ) ≃ α − ( v | w ) v . It then follows from Corollary 2.7 that either εd X ( x, x ′ ) < εd X ( x, x ′ )) ≃ d ε ( x, x ′ ) d ε ( x ) d ε ( x ′ ) . α − v | w ) v ˆ d ε ( x ) ˆ d ε ( x ′ ) ≤ α − ( n + m − d c X ( v,w )) α − n − α − m − ≃ α d c X ( v,w ) . Taking logarithms proves the first inequality in (6.2) of Definition 6.1.For the second inequality we distinguish two cases. Assume, without loss ofgenerality, that n ≤ m . If ˆ d ε ( x, x ′ ) > (1 − /α ) ˆ d ε ( x ), then α − n < α ˆ d ε ( x ) . ˆ d ε ( x, x ′ )and hence, by Lemma 3.3 and (6.4), α − ( v | w ) v ≃ ˆ d ε ( z, y ) + α − n + α − m < ˆ d ε ( x, x ′ ) + 3( α − n + α − m ) . ˆ d ε ( x, x ′ ) . This together with Corollary 2.7 implies that α d c X ( v,w ) = α − v | w ) v α − n α − m . ˆ d ε ( x, x ′ ) ˆ d ε ( x ) ˆ d ε ( x ′ ) . exp( εd X ( x, x ′ )) . Taking logarithms proves the second inequality in (6.2) in this case.If ˆ d ε ( x, x ′ ) ≤ (1 − /α ) ˆ d ε ( x ), then α − m ≥ ˆ d ε ( x ′ ) ≥ ˆ d ε ( x ) − ˆ d ε ( x, x ′ ) ≥ ˆ d ε ( x ) α > α − n − , and hence n + 1 ≥ m ≥ n . Let l and t be the smallest nonnegative integers suchthat α − l ≤ τ − ≤ α t . Then, by (6.4),ˆ d ε ( z, y ) ≤ ˆ d ε ( z, x ) + ˆ d ε ( x, x ′ ) + ˆ d ε ( x ′ , y ) < α − n + ˆ d ε ( x ) + 2 α − m ≤ α − n ≤ α t − n . If n ≥ t , then by Lemma 4.3 (with k = n − t ≥ d b X ( v, w ) ≤ n + m + 1 + 2 l − n − t ) ≤ l + t + 1) . If on the other hand n < t , then d b X ( v, w ) ≤ n + m ≤ n + 1 < t + 1 ≤ l + t + 1) . Thus, the second inequality in (6.2) holds also in the case ˆ d ε ( x, x ′ ) ≤ (1 − /α ) ˆ d ε ( x )by choosing C ≥ l + t + 1).To verify that some C ′ -neighborhood of Φ( X ) contains b X , note that every pointin b X is within a distance of the set V of vertices in b X . So it suffices to show thatif ( y, n ) ∈ V , then there is some x ∈ X such that d b X (( y, n ) , Φ( x )) ≤ C ′′ .As before, let l and t be the smallest nonnegative integers such that α − l ≤ τ − ≤ α t . Note that t ≥ 1. Recall the definition of S n from (6.3). Let x bea nearest point in S n + l + t to y . Then ˆ d ε ( y, x ) ≤ α − n − l − t e by Lemma 6.3. Bythe construction of Φ, the point x has a nearest point z n + l + t in A n + l + t such that( z n + l + t , n + l + t ) = Φ( x ) and ˆ d ε ( x, z n + l + t ) < α − n − l − t . For j = n, ... , n + l + t − z j ∈ A j such that ˆ d ε ( z, z j ) < α − j . Hence, by the choice of l and t ,ˆ d ε ( y, z n ) ≤ ˆ d ε ( y, x ) + ˆ d ε ( x, z ) + ˆ d ε ( z, z n ) < α − n − l − t e + 2 α − n − l − t + α − n < τ α − n . Since also y ∈ τ B Z ( y, α − n ) and z ∈ B Z ( z j , α − j ), j = n, ... , n + l + t , we see that( y, n ) ∼ ( z n , n ) ∼ ( z n +1 , n + 1) ∼ ... ∼ ( z n + l + t , n + l + t ) = Φ( x ) , where the first edge collapses into a single vertex if y = z n . This implies that d b X (( y, n ) , Φ( x )) ≤ l + 1. 7. Trees In this section, we will obtain a sharper version of Theorem 6.2 in the case when X is a tree, namely we get an isometry rather than a mere rough similarity, providedthat the parameters are chosen appropriately. Note that since X is a rooted tree, δ = 0 and we can choose ε (0) arbitrarily, by Theorem 2.5, so in this case there is noupper bound on ε in Theorem 6.2. Recall that an isometry is a 1-biLipschitz map,i.e. a rough similarity with L = 1 and C = 0. Theorem 7.1. Let X be a metric tree, rooted at x , such that every vertex x ∈ X has at least one child. Consider the uniformized boundary ∂ ε X of X , with parameter ε > , and let < τ < α = e ε be fixed. Let Z = ∂ ε X be equipped with the metric d Z ( ζ, ξ ) := ετ α d ε ( ζ, ξ ) . Then X is isometric to any hyperbolic filling of Z , constructed with the parameters α and τ . Note that diam Z ≤ τ /α < 1. In Remark 7.2 below we show that if τ ≥ α , thenthe hyperbolic filling is never a tree. Thus the range 1 < τ < α for τ in Theorem 7.1is optimal. Proof. Let b X be a hyperbolic filling of Z , constructed from a maximal α − n -separatedset A n ⊂ Z , with parameters τ and α . It suffices to show that the sets of verticesin X and b X , respectively, are isometric, since the extension to the edges is straight-forward.To start with, note that if ζ, ξ ∈ Z have a common ancestor x ∈ X at distance n ≥ x , then d Z ( ζ, ξ ) ≤ ετα Z ∞ n e − εt dt = τ α − n − < α − n , (7.1)with equality if and only if ζ and ξ do not have a common ancestor at distance n + 1from x . xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 23 It follows that for every n ≥ 0, the metric space Z can be written as a finite unionof open balls of radius α − n , namely those consisting exactly of all descendants in Z of some vertex x ∈ X with d X ( x, x ) = n . Moreover, every two points in such a ballsatisfy (7.1), and these balls are disjoint and can also be written as balls centeredat any of the points in it, with radius τ α − n . Indeed, if d Z ( ζ, η ) < τ α − n , then weknow by (7.1) and the comment after it that ζ and η have a common ancestor atdistance n from the root x , and so d Z ( ζ, η ) ≤ τ α − ( n +1) < α − n . Thus, A n containsexactly one point in each of these balls and this correspondence defines a bijection F between the vertices in X at level n and the set A n ⊂ Z . More precisely, F ( x ) isthe unique descendant of x belonging to A n . Define the mapping b F from verticesin X to vertices in b X as b F ( x ) = ( F ( x ) , d X ( x, x )) . To show that b F is an isometry between the two sets of vertices, assume that x ∼ y in X . Without loss of generality, we may assume that x is a parent of y and that d X ( x, x ) = n . Then both F ( x ) and F ( y ) have x as a common ancestor and henceby (7.1), d Z ( F ( x ) , F ( y )) < α − n , which yields F ( y ) ∈ B Z ( F ( x ) , α − n ) ∩ B Z ( F ( y ) , α − n − ) . Therefore b F ( x ) = ( F ( x ) , n ) ∼ ( F ( y ) , n + 1) = b F ( y ).Conversely, assume that b F ( x ) ∼ b F ( y ). Then | n − m | ≤ 1, where n = d X ( x, x )and m = d X ( y, x ). By the construction of b X , there exists ζ ∈ Z such that d Z ( ζ, F ( x )) < τ α − n and d Z ( ζ, F ( y )) < τ α − m . As pointed out after (7.1), the first inequality implies that ζ and F ( x ) have acommon ancestor at distance n from the root x and this ancestor must be x sincethere is only one ray in X from x to F ( x ). Similarly, ζ and F ( y ) have y as acommon ancestor at distance m from the root x .Because there is only one ray in X from x to ζ , this implies that x = y when m = n and contradicts the assumption b F ( x ) ∼ b F ( y ). Consequently, there are nohorizontal edges in b X . If m = n , then we can assume that m = n + 1 and concludethat x is the parent of y in the above ray, and so x ∼ y . Thus, b F : X → b X is anisometry. Remark 7.2. If Z is not a singleton in Theorem 7.1 (that is, X is not the half-line),then for τ ≥ α and a scaling of d Z so that diam Z < 1, the hyperbolic filling b X of Z will always contain horizontal edges and is thus not a tree.More precisely, let d Z ( ζ, ξ ) := εκd ε ( ζ, ξ ), where 0 < κ < 1, so that diam Z ≤ κ < 1. Let l ≥ κ < α − l . Then (7.1) becomes d Z ( ζ, ξ ) ≤ κα − n < α − n − l , and so Z can be written as a finite disjoint union ofopen balls of radius α − n − l , each consisting exactly of all descendants in Z of somevertex x ∈ X with d X ( x, x ) = n . Thus, A n + l contains exactly one point in each ofthese balls.However, if ζ, ξ ∈ A n + l are descendants of two distinct vertices x, y ∈ X atdistance n from the root, having the same parent, then because τ ≥ α , we have d Z ( ζ, ξ ) ≤ κα − n < α − n − l ≤ τ α − n − l . We therefore see that ζ ∈ τ B Z ( ζ, α − n − l ) ∩ τ B Z ( ξ, α − n − l ) and thus the vertices( ζ, n + l ) and ( ξ, n + l ) in b X will be neighbors connected by a horizontal edge.We also give the following characterizations. Proposition 7.3. Let X be a rooted tree such that every vertex x ∈ X has at leastone child, and let ε > . Then the following are true :(a) The uniformized boundary ∂ ε X is compact if and only if every vertex has afinite number of children. (b) The uniformized boundary ∂ ε X is doubling if and only if there is a uniformbound on the number of children for each vertex.Proof. Let 1 < τ < α = e ε and let b X be any hyperbolic filling of ∂ ε X , withparameters α and τ . By Theorem 7.1, b X is isometric to X . Part (a) now followsfrom Proposition 4.6, while part (b) follows from Proposition 4.5. 8. Geodesics in the hyperbolic filling In this section, except for Examples and , we fix an arbitrary parameter α > and assume that τ ≥ α + 1 α − . (8.1) We also assume that Z is a metric space with diam Z < , and let X be a hyperbolicfilling of Z with the parameters α and τ . In this section we study how the geodesics in a hyperbolic filling look like underthe above restriction relating τ and α . We do not use these precise properties ofgeodesics in the rest of the paper, and so in the other sections we do not require thislimit on τ . However, in other applications the structure of the geodesics is quiteuseful to know. As we gain control of the geodesics in a straightforward mannerunder the above constraint on τ , we have included this study here as well for theconvenience of the reader and for possible future applications. In Example 8.7 weshow that most of the properties obtained in this section can fail when τ is closeto 1. We end the section with Example 8.8 showing that when τ = 1 it is possiblefor the “hyperbolic filling” to be nonhyperbolic.Uniformizations will not play any role in this section. We will only studygeodesics between vertices in X and all the geodesics are with respect to the d X -metric. Lemma 8.1. Assume that (8.1) holds. If ( x, n ) ∼ ( z, n + 1) ∼ ( y, n ) is a segmentin a path with x = y , then ( x, n ) ∼ ( y, n ) and the path is not a geodesic.Proof. By the hypothesis of this lemma, we have B Z ( x, α − n ) ∩ B Z ( z, α − ( n +1) ) = ∅ and so, using (8.1), d Z ( x, z ) < α + 1 α α − n < τ α − n and similarly, d Z ( y, z ) < τ α − n . It follows that z ∈ τ B Z ( x, α − n ) ∩ τ B Z ( y, α − n ) and consequently ( x, n ) ∼ ( y, n ).So the length of the path can be reduced by one by replacing the segment ( x, n ) ∼ ( z, n + 1) ∼ ( y, n ) with the edge ( x, n ) ∼ ( y, n ). Hence it cannot be a geodesic. Lemma 8.2. Assume that (8.1) holds. If ( x , n ) ∼ ( x , n +1) ∼ ( y , n +1) ∼ ( y , n ) is a segment in a path with x = y , then ( x , n ) ∼ ( y , n ) and the path is not ageodesic.Proof. Since ( x , n + 1) ∼ ( y , n + 1), there exists b ∈ τ B Z ( x , α − ( n +1) ) ∩ τ B Z ( y , α − ( n +1) ) . xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 25 Similarly, as ( x , n ) ∼ ( x , n + 1), we have B Z ( x , α − n ) ∩ B Z ( x , α − ( n +1) ) = ∅ , andso d Z ( x , x ) < α + 1 α α − n . Therefore, d Z ( x , b ) ≤ d Z ( x , x ) + d Z ( x , b ) < α + 1 α α − n + τ α − ( n +1) ≤ τ α − n . Similarly, d Z ( y , b ) < τ α − n . Hence b ∈ τ B Z ( x , α − n ) ∩ τ B Z ( y , α − n ), which showsthat ( x , n ) ∼ ( y , n ).Finally, replacing the segment ( x , n ) ∼ ( x , n + 1) ∼ ( y , n + 1) ∼ ( y , n ) withthe edge ( x , n ) ∼ ( y , n ) reduces the length of the path by 2, and thus the originalpath is not a geodesic.Next, we show that there are no geodesics going first down (i.e. away from theroot) and then back up. The first part of this lemma will also be used when provingthe structure Lemma 8.6. Lemma 8.3. Assume that (8.1) holds. If ( x, n ) ∼ ( y, n +1) ∼ ( z, n +1) is a segmentin a geodesic, then there is some y ′ ∈ A n such that ( x, n ) ∼ ( y ′ , n ) ∼ ( z, n + 1) , andreplacing the segment ( x, n ) ∼ ( y, n + 1) ∼ ( z, n + 1) with ( x, n ) ∼ ( y ′ , n ) ∼ ( z, n + 1) also gives a geodesic.Consequently, if ( x , n ) ∼ ( x , n ) ∼ ... ∼ ( x m , n m ) is a geodesic, then thereare no indices ≤ i < j < k ≤ m with n j > max { n i , n k } .Proof. By the choice of A n there is some y ′ ∈ A n such that d Z ( z, y ′ ) < α − n . Weimmediately have that ( y ′ , n ) ∼ ( z, n + 1). Since we started with a geodesic, wesee that x = y ′ . Lemma 8.2 therefore implies that ( x, n ) ∼ ( y ′ , n ). Since ( x, n ) ∼ ( y, n +1) ∼ ( z, n +1) is a geodesic segment, it follows that ( x, n ) ∼ ( y ′ , n ) ∼ ( z, n +1)is also a geodesic segment, which proves the first claim.Next, assume that there would exist a geodesic violating the second part. Be-cause of Lemma 8.1, after restricting to a subpath we may assume that it is of theform ( x , n ) ∼ ( x , n + 1) ∼ ... ∼ ( x m − , n + 1) ∼ ( x m , n ) . (8.2)Applying the first part of the lemma iteratively shows that there are y , ... , y m − such that ( x , n ) ∼ ( y , n ) ∼ ... ∼ ( y m − , n ) ∼ ( x m − , n + 1) ∼ ( x m , n ) . As it has the same length as (8.2), it is also a geodesic, but this contradictsLemma 8.1. Lemma 8.4. Assume that (8.1) holds. Let v = ( x, n ) ∈ V . Then the following aretrue :(a) If v ∼ v ∼ ... ∼ v n = v and v ∼ w ∼ ... ∼ w n = v are two geodesics, then for j = 1 , ... , n − we have v j ∼ w j . (b) If ˆ v = ( y, n − ∼ v , then there is a geodesic v ∼ v ∼ ... ∼ v n = v such that v n − = ˆ v . In Example 8.7 below we show that (a) can fail drastically if τ is close to 1.Recall that here we assume that τ satisfies (8.1), and that v is the root of X .Part (b) holds for any hyperbolic filling, also without the requirement (8.1). Proof. To verify (a), note that by Lemma 3.1, v j = ( x j , j ) and w j = ( y j , j ) for some x j , y j ∈ A j , j = 0 , ... , n . Then w n − = ( y n − , n − ∼ v = ( x, n ) ∼ ( x n − , n − 1) = v n − and it follows from Lemma 8.1 that v n − ∼ w n − . Now an inductive application ofLemma 8.2 gives the desired conclusion.The second claim follows from the fact that the concatenation of the edge v ∼ ˆ v to any of the geodesics connecting ˆ v to the root vertex v gives a geodesic. Lemma 8.5. Assume that (8.1) holds. Let n be the smallest positive integer suchthat n α − n ≤ α + 1 . Then there is no horizontal geodesic of length ≥ n , i.e., if m ≥ n and ( y , n ) ∼ ( y , n ) ∼ ... ∼ ( y m , n ) , (8.3) then (8.3) is not a geodesic. If we drop the assumption (8.1) then the proof below shows that the same con-clusion holds provided that n is the smallest positive integer such that 2 n α − n ≤ − /τ . Proof. We may assume that m = 2 n . For 0 ≤ j ≤ n , there is some x j ∈ A j so that y ∈ B Z ( x j , α − j ). Necessarily, x = z and x n = y . Then for each j = 0 , ... , n − y ∈ B Z ( x j , α − j ) ∩ B Z ( x j +1 , α − ( j +1) ), and so( y , n ) ∼ ( x n − , n − ∼ ( x n − , n − ∼ ... ∼ ( x , ∼ v . (8.4)Similarly, we can find z j ∈ A j so that y m ∈ B Z ( z j , α − j ), 1 ≤ j < n , and( y m , n ) ∼ ( z n − , n − ∼ ( z n − , n − ∼ ... ∼ ( z , ∼ v . (8.5)If n ≤ n − 1, then combining (8.4) and (8.5) gives us a path from ( y , n ) to ( y m , n )(through the root v ) of length at most 2 n ≤ n − < m , and thus (8.3) is nota geodesic.Now assume that n ≥ n . Since d Z ( y j , y j +1 ) < τ α − n for 0 ≤ j ≤ n − , we get that d Z ( y , y n ) < n τ α − n . Let k = n + 1 − n . Then 1 ≤ k ≤ n and d Z ( x k , y n ) ≤ d Z ( x k , y ) + d Z ( y , y n ) < α − k + 2 n τ α − n = α − k (1 + 2 n τ α − n ) ≤ τ α − k , and in particular y n ∈ τ B Z ( x k , α − k ). Similarly, y n ∈ τ B Z ( z k , α − k ), and thus( x k , k ) ∼ ( z k , k ) . It follows that( y , n ) ∼ ( x n − , n − ∼ ... ∼ ( x k , k ) ∼ ( z k , k ) ∼ ... ∼ ( y n , n )is a path of length 2( n − k ) + 1 = 2 n − < n showing that (8.3) is not ageodesic.More general geodesics can be more complicated. However, we have the followinglemma, which can be used to obtain potentially simpler geodesics. xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 27 Lemma 8.6. Assume that (8.1) holds. If v = ( x, k ) and w = ( y, k m ) are twodistinct vertices with d X ( v, w ) = m , then there is a geodesic v = ( x , k ) ∼ ( x , k ) ∼ ... ∼ ( x m − , k m − ) ∼ ( x m , k m ) = w consisting of at most two vertical and one horizontal segments. More precisely, thereexist integers ≤ j ≤ j ≤ m , with j − j ≤ n − where n is as in Lemma ,such that k j +1 = k j − for ≤ j < j ,k j +1 = k j + 1 for j < j ≤ m,k j = k j = k j for j ≤ j ≤ j . This geodesic minimizes P j k j over all the geodesics between v and w , and hasa similar shape to the path identified in the latter part of Lemma 4.3. Proof. As X is connected, there is a geodesic between v and w . By the secondpart of Lemma 8.3 this geodesic does not contain any subpath going first down andthen up. We can therefore modify this geodesic iteratively using the first part ofLemma 8.3 to obtain a geodesic of the type described above. That j − j ≤ n − τ is small in comparison with α .The first example is tailored so that it can be used iteratively in the second ex-ample, producing a nonhyperbolic “hyperbolic filling” when τ = 1. This illustratesthe dependence of the Gromov constant δ on α and τ . Example 8.7. Let n ≥ α = 2, 1 ≤ τ < and ( τ − ≤ ρ < − n − with ρ > τ = 1. Set z = 0 , z = , z ± = ± ρ and Z = [ z , z − ] ∪ [ z + , z ] =: Z − ∪ Z + . Next we choose, A , A , ... , as follows: A = { z } , A = A = { z , z } ,A ′ j = { − j k : k = 0 , , ... , j − } ∩ Z, j = 3 , ... ,A j = A ′ j , j = 3 , ... , n − ,A j = ( A ′ j \ { z j ± } ) ∪ { z ± } , j = n, n + 1 , ... , where z j ± is the point in A ′ j closest to z ± . We then construct a “hyperbolic filling”based on this. In the first three levels we have( z , ∼ ( z , ∼ ( z , ∼ ( z , , ( z , ∼ ( z , ∼ ( z , 2) and ( z , ∼ ( z , ∼ ( z , . Next, for j = 2 , ... , n − 1, the distance between A j ∩ Z − and z + , as well as between A j ∩ Z + and z − , is | z j − − z + | = | z j + − z − | = 2 − j + ρ ≥ τ − j . Hence, there are no horizontal edges between the vertices ( z, j ) and ( y, i ) with z ∈ A j ∩ Z − , y ∈ A i ∩ Z + for 2 ≤ i, j ≤ n − . (8.6) On the other hand, since z + ∈ B Z ( z − , − n ) ∩ B Z ( z n − + , − n ), we see that( z − , n ) ∼ ( z n − + , n − 1) and similarly, ( z + , n ) ∼ ( z n − − , n − . (8.7)Hence, there are (at least) two upward-directed geodesics γ ± between ( z − , n ) and( z , γ − passing only through vertices with the first coordinate in Z − , whilethe vertices in γ + have the first coordinate in Z + , except for the starting and endingvertices. It follows that the midpoints in γ + and γ − have distance ( n − 1) to γ − and γ + , respectively, and so the Gromov constant δ ≥ ( n − n is even themidpoints of γ ± are not vertices.)With n = 3, this also shows that (at least) for τ < , it can happen that( z , ∼ ( z − , ∼ ( z , , while ( z , ( z , , i.e. both conclusions in Lemma 8.1, the last conclusion in Lemma 8.3, as well asLemma 8.4 (a) all fail in this case. Similarly, since( z , ∼ ( z + , ∼ ( z , ∼ ( z , y such that ( z , ∼ ( y, ∼ ( z , n = 4, the geodesics ( , ∼ ( z ± , ∼ ( , 3) are the only geodesicsbetween ( , 3) and ( , z , ∼ ( , ∼ ( z − , ∼ ( , Example 8.8. Let α = 2 and τ = 1. Let { n j } ∞ j =0 be an increasing sequence ofpositive integers n j ≥ N k = P kj =0 n j . Choose ρ j < − n j − and repeat theconstruction in Example 8.7 with n = n j and ρ = ρ j , and call the resulting space Z j , j = 0 , , ... .Now, replace the interval [0 , − N − ] ⊂ Z by a 2 − N -scaled copy of Z to formthe new space Z ′ = Z \ ( z ′ − , z ′ + ) , where z ′ ± := 2 − N ( ± ρ ) . The first two points z and 2 − N − in A N +1 ⊂ Z ′ are still next to each other andthe corresponding vertices form the horizontal edge( z , N + 1) ∼ (2 − N − , N + 1)similarly to ( z , ∼ ( z , 1) in Z .On the other hand, in the following levels j = N +2 , ... , N − 1, similarly to (8.6),there are no horizontal edges between the left-most interval [0 , z ′ − ] and the rest of Z ′ . At the same time, similarly to (8.7), the points z ′ ± ∈ A N have upward-directededges both to the interval [0 , z ′ − ] and the second interval in Z ′ . Consequently, thevertex ( z ′ − , N ) has two upward-directed geodesics γ ′ ± to ( z , N + 1), such that themidpoints of γ ′ + and γ ′ − have distance ( n − 1) to γ ′ − and γ ′ + , respectively.Next, the interval [0 , − N − ] ⊂ Z ′ can be replaced by a 2 − N -scaled copy of Z , i.e. we get the new space Z ′ = Z ′ \ (2 − N ( − ρ ) , − N ( + ρ )). Continuingin this way, we obtain a compact doubling space Z ′ = ∞ \ j =1 Z ′ j . Moreover, if z ∈ Z ′ and 0 < r < , then m ( B Z ( z, r ) ∩ Z ′ ) ≃ m ( B Z ( z, r )), where m denotes the Lebesgue measure.Since lim j →∞ n j = ∞ , we can for each k find two vertices having two upward-directed geodesics e γ ± between them such that the midpoint of e γ + has distance ≥ k to e γ − , i.e. the hyperbolic filling of Z ′ does not satisfy the Gromov δ -condition, andis thus not Gromov hyperbolic. xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 29 9. Measures, function spaces and capacities In this section, we assume that ≤ p < ∞ and that ( Y, d ) is a metric space equippedwith a complete Borel measure ν such that < ν ( B ) < ∞ for all balls B ⊂ Y . Wecall Y = ( Y, d, ν ) a metric measure space. In the rest of the paper we are interested in studying the metric space Z , con-sidered in the previous sections, together with a doubling measure on Z and Besovspaces on Z with respect to this measure. In particular, as mentioned in the in-troduction, we wish to associate Besov functions on Z with upper gradient-basedSobolev functions on the uniformization X ε of the hyperbolic filling X of Z . In thissection we will explain the notions related to measures and function spaces.We follow Heinonen–Koskela [26] in introducing upper gradients as follows (theyare referred to as very weak gradients in [26]). For proofs of the facts on uppergradients and Newtonian functions discussed in this section, we refer the reader toBj¨orn–Bj¨orn [2] and Heinonen–Koskela–Shanmugalingam–Tyson [27]. Definition 9.1. A Borel function g : Y → [0 , ∞ ] is an upper gradient of a function u : Y → [ −∞ , ∞ ] if for each nonconstant compact rectifiable curve γ in Y , we have | u ( x ) − u ( y ) | ≤ Z γ g ds. (9.1)Here x and y denote the two endpoints of γ . The above inequality should beinterpreted as also requiring that R γ g ds = ∞ if at least one of u ( x ) and u ( y ) isnot finite. If g is a nonnegative measurable function on Y and if (9.1) holds for p -almost every curve (see below), then g is a p -weak upper gradient of u .We say that a property holds for p -almost every curve if the family Γ of allnonconstant compact rectifiable curves for which the property fails has zero p -modulus , i.e. there is a Borel function 0 ≤ ρ ∈ L p ( Y ) such that R γ ρ ds = ∞ for every curve γ ∈ Γ. The p -weak upper gradients were introduced in Koskela–MacManus [33]. It was also shown therein that if g ∈ L p ( Y ) is a p -weak uppergradient of u , then one can find a sequence { g j } ∞ j =1 of upper gradients of u suchthat k g j − g k L p ( Y ) → u has an upper gradient in L p ( Y ), then it has a minimal p -weak upper gradient g u ∈ L p ( Y ) in the sense that g u ≤ g a.e. for every p -weak upper gradient g ∈ L p ( Y )of u , see Shanmugalingam [46]. The minimal p -weak upper gradient is well definedup to a set of measure zero.Following Shanmugalingam [45], we define a version of Sobolev spaces on Y . Definition 9.2. A function u : Y → [ −∞ , ∞ ] is in the Newtonian space e N ,p ( Y )if R Y | u | p dµ < ∞ and u has a p -weak upper gradient g ∈ L p ( Y ). This space is avector space and a lattice, equipped with the seminorm k u k N ,p ( Y ) given by k u k N ,p ( Y ) := (cid:18)Z Y | u | p dν + inf g Z Y g p dν (cid:19) /p , where the infimum is taken over all upper gradients g of u , or equivalently all p -weakupper gradients g of u (see the comments above).The Newtonian space N ,p ( Y ) = e N ,p ( Y ) / ∼ , where ∼ is the equivalence relationon e N ,p ( Y ) given by u ∼ v if and only if k u − v k N ,p ( Y ) = 0. To specify the measurewith respect to which the Newtonian space is taken, we will also write e N ,p ( Y, ν )and N ,p ( Y, ν ). Definition 9.3. The C Yp -capacity of a set E ⊂ Y is defined as C Yp ( E ) = inf u k u k pN ,p ( Y ) , where the infimum is taken over all u ∈ e N ,p ( Y ) satisfying u ≥ E .Note that since functions in e N ,p ( Y ) are defined pointwise everywhere, the re-quirement u ≥ E in the definition of C Yp ( E ) makes sense for an arbitrary set E ⊂ Y .A property is said to hold quasieverywhere (q.e. or C Yp -q.e.) if the set of allpoints at which the property fails has C Yp -capacity zero. The capacity is the correctgauge for distinguishing between two Newtonian functions. If u ∈ e N ,p ( Y ), then v ∼ u if and only if v = u q.e. Moreover, if u, v ∈ e N ,p ( Y ) and u = v a.e., then u = v q.e. That means that the equivalence classes in N ,p ( Y ) are precisely madeup of functions which are equal q.e., and not a.e. as in the usual Sobolev spaces. Byan abuse of notation, just as for L p -spaces, we will often not distinguish between afunction in e N ,p ( Y ) and the corresponding equivalence class in N ,p ( Y ). Definition 9.4. We say that Y (or the measure ν ) supports a p -Poincar´e inequality if there exist C, λ > B = B ( x, r ) and for all integrablefunctions u and upper gradients g of u on λB , Z B | u − u B | dν ≤ Cr (cid:18)Z λB g p dν (cid:19) /p , where u B := R B u dν = ν ( B ) − R B u dν .See Bj¨orn–Bj¨orn [2] and Heinonen–Koskela–Shanmugalingam–Tyson [27] forequivalent formulations of the C Yp capacity and the p -Poincar´e inequality. Remark 9.5. We will primarily be interested in Newtonian spaces on the uni-formization X ε of a hyperbolic filling of Z , and on its closure X ε , in both casesequipped with the measure µ β , β > 0, defined by (10.7) below. In particular, eachedge in X is measured by a multiple of the Lebesgue measure. It is then quite easyto see that the only family of nonconstant compact rectifiable curves in X ε which haszero p -modulus (with respect to µ β ) is the empty family. Functions in Newtonianspaces are absolutely continuous on p -almost every line, see Shanmugalingam [45].Thus all functions u ∈ e N ,p ( X ε , µ β ) are continuous on X ε and absolutely continu-ous on each edge. Moreover, g u = | du/ds ε | a.e. on each edge, where ds ε denotes thearc length with respect to d ε . In particular, each equivalence class in N ,p ( X ε , µ β )contains just one function, and that function is continuous. Moreover, points in X ε have positive capacity.For functions on X ε , the situation is not quite as simple, but the following resultwill be useful. A function u on Y is C Yp -quasicontinuous if for each η > G ⊂ Y with C Yp ( G ) < η such that u | Y \ G is continuous. Note that any E ⊂ X ε with C X ε p ( E ) = 0 must satisfy E ⊂ ∂ ε X . Theorem 9.6. (Bj¨orn–Bj¨orn–Shanmugalingam [4]) Assume that Y is complete andthat ν is doubling and supports a p -Poincar´e inequality. Then every u ∈ e N ,p ( Y ) is C Yp -quasicontinuous. Moreover, C Yp is an outer capacity, i.e. C Yp ( E ) = inf G ⊃ EG open C Yp ( G ) . xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 31 We will use these facts together with our trace and extension results to show thatBesov functions on Z have Cap B θp,p ( Z ) -quasicontinuous representatives (which isdefined just as C Yp -quasicontinuous), see Proposition 13.3. Similarly, we will obtaindensity of Lipschitz functions and existence of H¨older continuous representatives inBesov spaces using our trace and extension results, and corresponding theorems forNewtonian functions.We now give the definition of Besov spaces on metric measure spaces. Definition 9.7. Let θ > 0. We say that u ∈ L p ( Y ) is in the Besov space B θp,p ( Y ) if k u k pθ,p := Z Y Z Y | u ( ζ ) − u ( ξ ) | p d ( ζ, ξ ) pθ dν ( ξ ) dν ( ζ ) ν ( B ( ζ, d ( ζ, ξ ))) < ∞ . Remark 9.8. Note that B θp,p ( Y ) is a Banach space with the norm given by k u k B θp,p ( Y ) = k u k θ,p + k u k L p ( Y ) . Indeed, it is clear that this function space is a normed vector space. To see thatit is complete, we argue as follows. Let { u k } ∞ k =1 be a Cauchy sequence in B θp,p ( Y ).Then it is a Cauchy sequence in L p ( Y ), and hence it is convergent to some function u ∈ L p ( Y ). By passing to a subsequence if necessary, we also ensure that u k → uν -a.e. in Y . Setting the measure ν on Y × Y by ν ( E ) = Z E d ( ν × ν )( ξ, ζ ) d ( ζ, ξ ) pθ ν ( B ( ζ, d ( ζ, ξ ))) , and defining v k : Y × Y → R as v k ( ξ, ζ ) = u k ( ξ ) − u k ( ζ ), we note that k u k k θ,p = k v k k L p ( Y × Y,ν ) . Thus, { v k } ∞ k =1 is also a Cauchy sequence in the complete space L p ( Y × Y, ν ), andso converges therein to a function v : Y × Y → R . Again, by passing to yet anothersubsequence if necessary, we may also assume that v k → v ν -a.e. in Y × Y . Setting w : Y × Y → R by w ( ξ, ζ ) = u ( ξ ) − u ( ζ ), we see that necessarily v = w ν -a.e. in Y × Y . Therefore v k → w in L p ( Y × Y, ν ), that is, u k → u in B θp,p ( Y ).We recall the following lemma. For a proof see Gogatishvili–Koskela–Shanmu-galingam [21, Theorem 5.2 and (5.1)] (where the factor of 2 should be replaced with α > Lemma 9.9. Assume that ν is doubling and θ > . If u ∈ B θp,p ( Y ) , then k u k pθ,p ≃ ∞ X n =0 Z Y Z B ( ζ,α − n ) | u ( ζ ) − u ( η ) | p α − nθp dν ( ζ ) dν ( η ) . Definition 9.10. We set the Besov capacity of E ⊂ Y to be the numberCap B θp,p ( Y ) ( E ) := inf u ( k u k pθ,p + k u k pL p ( Y ) ) , where the infimum is taken over all u ∈ B θp,p ( Y ) satisfying u ≥ E . 10. Lifting doubling measures from Z to its hyper-bolic filling X From now on, we let X be a hyperbolic filling, constructed with parameters α, τ > ,of a compact metric space Z with < diam Z < and equipped with a doubling measure ν . In this section, we also let X ε be the uniformization of X with parameter < ε ≤ log α . We now focus on lifting up ν from Z as follows: Recall that the vertices in X are denoted v = ( z, n ), where z belongs to the maximal α − n -separated set A n ⊂ Z .Note that n is the graph distance from the root v := ( z , 0) to ( z, n ). For ( z, n ) ∈ V we set ˆ µ ( { ( z, n ) } ) = ν ( B Z ( z, α − n )) . (10.1)The measure µ on X is then given by “smearing out” ˆ µ to X : for a Borel set A ⊂ X , µ ( A ) = X v ∈ V X w ∼ v (cid:0) ˆ µ ( { v } ) + ˆ µ ( { w } ) (cid:1) L ( A ∩ [ v, w ]) , (10.2)where [ v, w ] denotes the unit interval that connects the two vertices v and w , and L denotes the Lebesgue measure.Strictly speaking, it is ˆ µ ( v ) deg v that is smeared out, but because X has uni-formly bounded degree (by Proposition 4.5) this is comparable to ˆ µ ( v ).Note that the vertex set V n of points in X that are at level n from the root iscomposed of a maximal α − n -separated set of points from Z . So by the work of Gilland Lopez [20], [36], we know that V n , equipped with the neighborhood relationshipinherited from V and with the measure ˆ µ | V n , is doubling and that a subsequenceconverges in the pointed measured Gromov–Hausdorff sense to a measure ˆ µ ∞ on Z such that ˆ µ ∞ ≃ ν . Lemma 10.1. Let ( z, n ) , ( y, m ) ∈ V with ( y, m ) ∼ ( z, n ) . Then C Nd ˆ µ ( { ( z, n ) } ) ≤ ˆ µ ( { ( y, m ) } ) ≤ C Nd ˆ µ ( { ( z, n ) } ) , where N is the smallest integer such that N ≥ α (1 + τ ) + τ and C d is the doublingconstant associated with ν .Proof. Since the two points are neighbors, we have that | n − m | ≤ α − n ≤ α − m . Since τ B Z ( z, α − n ) ∩ τ B Z ( y, α − m ) = ∅ by the construction of thehyperbolic filling, every ζ ∈ B Z ( z, α − n ) satisfies d Z ( ζ, y ) ≤ d Z ( ζ, z ) + d Z ( z, y ) < α − n + τ ( α − n + α − m ) ≤ N α − m , and so B Z ( z, α − n ) ⊂ N B Z ( y, α − m ). The doubling property of ν then implies thatˆ µ ( { ( z, n ) } ) = ν ( B Z ( z, α − n )) ≤ C Nd ν ( B Z ( y, α − m )) = C Nd ˆ µ ( { ( y, m ) } ) . Reversing the roles of z and y in the above argument gives the desired doubleinequality. Theorem 10.2. Let Y be a metric graph equipped with the length metric d suchthat all edges have length and assume that Y has uniformly bounded degree, i.e.every vertex has at most K neighbors. Let ˆ µ be a discrete measure defined on thevertices of Y and such that ˆ µ ( v ) ≃ ˆ µ ( w ) whenever v ∼ w , with comparison constantsindependent of v and w . Consider the smeared out measure µ on Y given by (10.2) .Then for each R > there is a constant C ≥ such that for all balls B = B ( x, r ) with r ≤ R , and every integrable function u and upper gradient g of u on B , µ (2 B ) ≤ C µ ( B ) and Z B | u − u B | dµ ≤ C r Z B g dµ. (10.3) The constant C depends only on R , K and the comparison constants in ˆ µ ( v ) ≃ ˆ µ ( w ) . xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 33 Proof. Since Y is a length space, it follows from Lemma 3.5 and Theorem 5.3 inBj¨orn–Bj¨orn–Shanmugalingam [5] that it suffices to consider only the case R = .Let v be a nearest vertex to the center x of B , i.e. d ( x, v ) = dist( x, V ). As r ≤ ,the ball 2 B contains at most one vertex, namely v . Hence, B ⊂ B ⊂ [ w ∼ v [ v, w ] and dµ = (ˆ µ ( { v } ) + ˆ µ ( { w } )) ds ≃ ˆ µ ( { v } ) ds on each [ v, w ] , (10.4)by Lemma 10.1. Thus µ (2 B ) . Kr ˆ µ ( { v } ) . µ ( B ) . To prove the 1-Poincar´e inequality in (10.3), observe that if v / ∈ B , then B is an interval, and so the 1-Poincar´e inequality for B follows from the 1-Poincar´einequality on R and the fact that dµ = C B d L on B . On the other hand, if v ∈ B ,then B = [ w ∼ v I w , where I w = B ∩ [ v, w ] . We therefore obtain from (10.4) and the definition of upper gradients that for each w ∼ v , Z I w | u − u ( v ) | dµ ≤ Z I w Z I w g ( s ) ds dµ = µ ( I w ) Z I w g ds . r ˆ µ ( { v } ) Z I w g ds ≃ r Z I w g dµ. Summing over all w ∼ v yields Z B | u − u ( v ) | dµ = X w ∼ v Z I w | u − u ( v ) | dµ . X w ∼ v r Z I w g dµ = r Z B g dµ. A standard argument based on the triangle inequality allows us to replace u ( v ) onthe left-hand side by u B at the cost of an extra factor 2 on the right-hand side.To obtain a doubling measure on X ε with respect to the uniformized metric d ε ,we can equip X ε with the uniformized measure d ˆ µ β ( x ) = ρ β ( x ) dµ ( x ) , where ρ β ( x ) = e − βd X ( x,v ) ≃ d ε ( x ) β/ε . (10.5)Theorems 4.9 and 6.2 in Bj¨orn–Bj¨orn–Shanmugalingam [5] (which hold for generalGromov hyperbolic spaces) then guarantee that for sufficiently large β , the obtainedmeasure ˆ µ β is doubling and supports a 1-Poincar´e inequality on X ε as well as on X ε .More precisely, this holds for β > β , where β is determined by the limitations givenin [5, Theorems 4.9 and 6.2] based on the doubling constant from Theorem 10.2.In considering the specific case of hyperbolic fillings, we are able to show thatthe requirement β > β can be omitted. Just as we showed in Theorem 5.1 thatthe full range 0 < ε ≤ log α is possible, we can allow for the full range β > Z , it is possible to start directly with the weighted discretemeasure ρ β ( v )ˆ µ ( { v } ) = e − βn ν ( B Z ( z, α − n )) for v = ( z, n ) ∈ V, (10.6)defined on the vertices in X , and smear it out as in (10.2): µ β ( A ) = X v ∈ V X w ∼ v (cid:0) ρ β ( v )ˆ µ ( { v } ) + ρ β ( w )ˆ µ ( { w } ) (cid:1) L ( A ∩ [ v, w ]) . (10.7)Since ρ β ( v ) ≃ ρ β ( w ) whenever v ∼ w , the measures µ β and ˆ µ β are clearly compa-rable, and the following result also holds for ˆ µ β . Theorem 10.3. For every β > , µ β is doubling and supports a -Poincar´e in-equality on X ε as well as on X ε .Furthermore, for all ζ ∈ ∂ ε X and < r ≤ ε X ε ≤ /ε , µ β ( B ε ( ζ, r )) ≃ ( εr ) β/ε ν ( B Z ( ζ, ( εr ) /σ )) , where σ = ε log α ≤ and the comparison constants depend only on ε , β , α , τ and the doubling constantassociated with ν . In particular, ν is comparable to the β/ε -codimensional measure on ∂ ε X gener-ated by µ β . Lemma 10.4. Let < ε ≤ log α . Then B X (cid:18) x, C rεd ε ( x ) (cid:19) ⊂ B ε ( x, r ) ⊂ B X (cid:18) x, C rεd ε ( x ) (cid:19) , if x ∈ X and < r ≤ d ε ( x ) , where C , C > are independent of ε .Proof. This was proved when 0 < ε ≤ ε ( δ ) for general Gromov hyperbolic spaces(where δ is the Gromov hyperbolicity constant) in Theorem 2.1 in Bj¨orn–Bj¨orn–Shanmugalingam [5], with Remark 2.11 in [5] showing that C , C > ε . That proof only relies on the following facts which hold for hyperbolicfillings for all 0 < ε ≤ log α : • That X is geodesic, which follows from the definition. • That X is locally compact and X ε is geodesic, which follows from Proposi-tion 4.6, as Z is equipped with a doubling measure. • Lemma 2.6 (i.e. [8, Lemma 4.16]) which holds for arbitrary ε > Proof of Theorem . We first concentrate on the doubling property. We needto consider three types of balls, namely, subWhitney balls, balls centered in (ornear) ∂ ε X and intermediate balls. Recall that the graph X has uniformly boundeddegree, by Proposition 4.5.1. For subWhitney balls , that is, balls B ε ( x, r ) with r ≤ d ε ( x ), we note that2 C r/εd ε ( x ) ≤ C / ε . Hence, Lemmas 10.1, 10.4 and Theorem 10.2 imply that µ β ( B ε ( x, r )) ≤ µ β (cid:18) B X (cid:18) x, C rεd ε ( x ) (cid:19)(cid:19) ≃ ρ β ( x ) µ (cid:18) B X (cid:18) x, C rεd ε ( x ) (cid:19)(cid:19) ≃ µ β (cid:18) B X (cid:18) x, C rεd ε ( x ) (cid:19)(cid:19) ≤ µ β ( B ε ( x, r )) . 2. If x ∈ X ε and r ≥ d ε ( x ), then for some ζ ∈ ∂ ε X , B ε ( x, r ) ⊂ B ε ( ζ, r ) and B ε ( ζ, r ) ⊂ B ε ( x, r ) . It therefore suffices to estimate µ β ( B ε ( ζ, r )) for all ζ ∈ ∂ ε X and 0 < r ≤ ε X ε .From the construction of µ β , and using also the uniformly bounded degree of X , itis clear that µ β ( B ε ( ζ, r )) ≃ X v ∈ V ∩ B ε ( ζ,r ) e − βπ ( v ) ˆ µ ( v ) . Let v = ( z, n ) ∈ V ∩ B ε ( ζ, r ). Then e − εn = εd ε ( v ) < εr by (4.1), and hence n ≥ N ,where N is the smallest nonnegative integer such that N ≥ ε log 1 εr . (10.9) xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 35 Let z j ∈ A j ⊂ Z be such that d Z ( z j , ζ ) < α − j , j = 0 , , ... . Proposition 4.4 showsthat d Z ( z, z j ) σ ≤ (2 τ α ) σ d ε ( v, ( z j , j )) ≤ (2 τ α ) σ (cid:0) d ε ( v, ζ ) + d ε ( ζ, ( z j , j )) (cid:1) . Since the path ( z , ∼ ... ∼ ( z j , j ) ∼ ... is a geodesic ray in X ending at ζ (seeLemma 3.1 and Proposition 4.4), we have d ε ( ζ, ( z j , j )) = e − εj /ε. (10.10)Letting j → ∞ then shows that d Z ( z, ζ ) ≤ d Z ( z, z j ) + d Z ( z j , ζ ) ≤ τ α (cid:18) r + e − εj ε (cid:19) /σ + α − j → τ αr /σ . We therefore obtain that µ β ( B ε ( ζ, r )) . X n ≥ N e − βn X z ∈ A n ∩ B Z ( ζ, ταr /σ ) ν ( B Z ( z, α − n )) . Since α − n = ( e − εn ) /σ < ( εr ) /σ , the bounded overlap of the balls B Z ( z, α − n ) in A n ⊂ Z and the doubling property of ν now imply that µ β ( B ε ( ζ, r )) . X n ≥ N e − βn ν ( B Z ( ζ, ( εr ) /σ )) ≃ ( εr ) β/ε ν ( B Z ( ζ, ( εr ) /σ )) . (10.11)To verify the reverse comparison, observe that by (10.9) and (10.10) we have d ε ( ζ, ( z N , N )) = e − εN /ε < r . It follows that the edge ( z N , N ) ∼ ( z N +1 , N + 1) iscontained in B ε ( ζ, r ). Consequently, using the doubling property of ν and the factthat d Z ( ζ, z N ) < α − N = ( e − εN ) /σ ≃ ( εr ) /σ , we conclude from (10.7) that µ β ( B ε ( ζ, r )) & e − βN ˆ µ ( { ( z N , N ) } ) ≃ ( εr ) β/ε ν ( B Z ( z N , α − N )) ≃ ( εr ) β/ε ν ( B Z ( ζ, ( εr ) /σ )) , which, together with (10.11), proves (10.8).3. If x ∈ X ε and d ε ( x ) ≤ r ≤ d ε ( x ), then clearly B ε ( x, d ε ( x )) ⊂ B ε ( x, r ) ⊂ B ε ( x, r ) ⊂ B ε ( x, d ε ( x )) . From Proposition 4.6 we know that X ε is compact, and thus there is ζ ∈ ∂ ε X suchthat d ε ( ζ, x ) = d ε ( x ). Let γ be a d ε -geodesic from x to ζ and let v = ( z, k ) be thevertex in γ nearest to x . As in case 1, using (10.7), Lemma 10.1, and the uniformboundedness of the degrees, we see that µ β ( B ε ( x, d ε ( x ))) ≃ ρ β ( x )ˆ µ ( { v } ) = ρ β ( x ) ν ( B Z ( z, α − k )) . (10.12)On the other hand, by (10.8) (proved when considering case 2 above), µ β ( B ε ( x, d ε ( x ))) ≤ µ β ( B ε ( ζ, d ε ( x ))) ≃ ( εd ε ( x )) β/ε ν ( B Z ( ζ, (5 εd ε ( x )) /σ )) . Note that the right-hand side of the above is comparable to the right-hand side of(10.12) since d ε ( z, ζ ) ≤ d ε ( z, v ) + d ε ( v, ζ ) ≤ d ε ( x ) , ν is doubling, and( εd ε ( x )) β/ε ≃ ρ ε ( x ) β/ε = ρ β ( x ) and α − k = e − εk/σ ≃ ρ ε ( x ) /σ ≃ ( εd ε ( x )) /σ . The doubling property of µ β now follows directly in all three cases from the aboveestimates. To show that µ β supports a 1-Poincar´e inequality on X ε , we proceed as in theproof of Lemma 6.1 in Bj¨orn–Bj¨orn–Shanmugalingam [5]. This is possible for all β > 0, not only β > β as in [5], since we already know that µ β is doubling on X ε . Together with Theorem 10.2, it shows that there exists c > B = B ε ( x, r ) with x ∈ X , the measure µ β , and0 < r ≤ c d ε ( x ). Since X ε is a uniform length space and µ β is doubling on X ε , weconclude from [5, Proposition 6.3] that µ β supports a 1-Poincar´e inequality on X ε as well as on X ε . Corollary 10.5. With the assumptions as in Theorem , we have for all x ∈ X ε and < r ≤ ε X ε , µ β ( B ε ( x, r )) ≃ ( ( εr ) β/ε ν ( B Z ( ζ, ( εr ) /σ )) , if r ≥ d ε ( x ) and ζ ∈ Z is a nearest point to x,r ( εd ε ( x )) β/ε − ˆ µ ( { v } ) , if r ≤ d ε ( x ) and v ∈ X is a nearest vertex to x. In both cases, the nearness is with respect to the metric d ε . Recall from (4.1) that α − n = ( e − εn ) /σ = ( εd ε ( v )) /σ if v = ( z, n ) ∈ V . More-over, if ζ ∈ Z and v = ( z, n ) ∈ V are nearest points to x in Z and V , respectively,then by Proposition 4.4, d Z ( ζ, z ) σ ≃ d ε ( ζ, z ) . d ε ( x ) ≃ d ε ( v ) . It therefore follows from (10.1) and the doubling property of ν thatˆ µ ( { v } ) = ν ( B Z ( z, ( εd ε ( v )) /σ ) ≃ ν ( B Z ( ζ, ( εd ε ( x )) /σ )) , which further simplifies the formula in Corollary 10.5. Also note that since ν is dou-bling, B Z ( ζ, ( εd ε ( x )) /σ )) can be replaced by any ball B Z ( ξ, ( εd ε ( x )) /σ )) with ξ ∈ Z such that d ε ( ξ, x ) . d ε ( x ), and that ν ( B Z ( ζ, ( εd ε ( x )) /σ )) ≃ ν ( B ε ( ζ, εd ε ( x ))). Proof of Corollary . The first case follows directly from (10.8) together withthe inclusions B ε ( ζ, r ) ⊂ B ε ( x, r ) ⊂ B ε ( ζ, r )and the doubling property of µ β and ν .In the second case, Lemma 10.4 implies that B ε ( x, r ) ⊂ B X (cid:18) x, C r εd ε ( x ) (cid:19) ⊂ B X (cid:18) x, C ε (cid:19) Recall from Proposition 4.5 that the graph X has uniformly bounded degree. There-fore we have by (10.7) and Lemma 10.1 that µ β ( B ε ( x, r )) ≃ rεd ε ( x ) (cid:18) ρ β ( v )ˆ µ ( { v } ) + X w ∈ V ∩ B ε ( x,r/ ρ β ( w )ˆ µ ( { w } ) (cid:19) ≃ rεd ε ( x ) ρ β ( v )ˆ µ ( { v } ) . Since ρ β ( v ) ≃ ( εd ε ( x )) β/ε , the doubling property of µ β concludes the proof. Lemma 10.6. Assume that the measure ν on Z satisfies for all ζ ∈ Z and < r ′ ≤ r ≤ Z , ν ( B Z ( ζ, r ′ )) ν ( B Z ( ζ, r )) & (cid:18) r ′ r (cid:19) s ν . (10.13) xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 37 Let β > and ε = log α . Then the measure µ β , defined by (10.6) and (10.7) ,satisfies for all x ∈ X ε and < r ′ ≤ r ≤ ε X ε , µ β ( B ε ( x, r ′ )) µ β ( B ε ( x, r )) & (cid:18) r ′ r (cid:19) s β , where s β = max { , β/ε + s ν } . (10.14)It is well known that every doubling measure ν satisfies (10.13) for some s ν > Proof. Note that σ = 1. We shall distinguish three cases:1. If r ≤ d ε ( x ) then the second case in Corollary 10.5 applies both to r and r ′ and hence µ β ( B ε ( x, r ′ )) µ β ( B ε ( x, r )) ≃ r ′ r . 2. If r ′ ≥ d ε ( x ) then the first case in Corollary 10.5 applies both to r and r ′ andhence µ β ( B ε ( x, r ′ )) µ β ( B ε ( x, r )) ≃ ( εr ′ ) β/ε ν ( B Z ( ζ, εr ′ ))( εr ) β/ε ν ( B Z ( ζ, εr )) & (cid:18) r ′ r (cid:19) β/ε + s ν . 3. If r ′ ≤ d ε ( x ) ≤ r then by the already proved cases 1 and 2, µ β ( B ε ( x, r ′ )) µ β ( B ε ( x, r )) = µ β ( B ε ( x, r ′ )) µ β ( B ε ( x, d ε ( x ))) µ β ( B ε ( x, d ε ( x ))) µ β ( B ε ( x, r )) & r ′ d ε ( x ) (cid:18) d ε ( x ) r (cid:19) β/ε + s ν . If β/ε + s ν ≥ r ′ d ε ( x ) (cid:18) d ε ( x ) r (cid:19) β/ε + s ν ≥ (cid:18) r ′ d ε ( x ) (cid:19) β/ε + s ν (cid:18) d ε ( x ) r (cid:19) β/ε + s ν = (cid:18) r ′ r (cid:19) β/ε + s ν , and if β/ε + s ν ≤ 1, then r ′ d ε ( x ) (cid:18) d ε ( x ) r (cid:19) β/ε + s ν ≥ r ′ d ε ( x ) d ε ( x ) r = r ′ r . From the above three cases we conclude that (10.14) holds. 11. Traces to Z from the hyperbolic filling X Recall the standing assumptions from Section . Here and in the rest of the paper,we also let ≤ p < ∞ and consider the uniformized space X ε equipped with themeasure µ β where ε = log α and β > . Theorem 5.1 shows that X ε is a uniform space. From Proposition 4.4 with ε =log α we know that ∂ ε X is biLipschitz equivalent to Z . Hence we can replace ∂ ε X by Z as well, since the Besov spaces are biLipschitz invariant. Of course, the measureon Z is pushed forward to ∂ ε X via the biLipschitz identification Ψ : Z → ∂ ε X . Inthe following, we shall therefore not distinguish between ( ∂ ε X, d ε ) and ( Z, d Z ).We equip the uniformized space X ε with the doubling measure µ β , obtained in(10.7). Equivalently, the uniformized measure ˆ µ β from (10.5), based on the smearedout measure µ from (10.2), can be used.For the vertices in X , consider the projections π (( z, n )) = z and π (( z, n )) = n .Whenever a nonvertex x ∈ X belongs to the edge [ v, w ] ⊂ X , let π ( x ) := min { π ( v ) , π ( w ) } . (11.1) Theorem 11.1. Let u ∈ e N ,p ( X ε , µ β ) and < θ ≤ − β/εp . Then u has a trace ˜ u ∈ B θp,p ( Z ) given by (11.4) and (11.5) below, with the ( semi ) norm estimates k ˜ u k θ,p . k g u k L p ( X ε ,µ β ) (11.2) and k ˜ u k L p ( Z ) . | u ( v ) | + k g u k L p ( X ε ,µ β ) . k u k N ,p ( X ε ,µ β ) . (11.3) If u ∈ Lip( X ε ) then ˜ u = ˆ u | Z , where ˆ u is the unique Lipschitz extension of u to X ε . Note that the equivalence classes in N ,p ( X ε , µ β ) consist of one function each,see Remark 9.5. Proof. Let u ∈ e N ,p ( X ε , µ β ) with an upper gradient g ∈ L p ( X ε , µ β ). For ζ ∈ Z and n = 0 , , ... , let A n ( ζ ) = A n ∩ B Z ( ζ, α − n ). We define u n ( ζ ) = 1 A n ( ζ ) X z ∈ A n ( ζ ) u (( z, n )) , (11.4)where A n ( ζ ) is the cardinality of A n ( ζ ). Note that the construction of A n , to-gether with the doubling property of Z , shows that 1 ≤ A n ( ζ ) ≤ K for some K independent of n and ζ .For each fixed z ∈ A n , the function χ B Z ( z,α − n ) is lower semicontinuous and thus ν -measurable. Hence also the linear combinations X z ∈ A n ( ζ ) u (( z, n )) = X z ∈ A n u (( z, n )) χ B Z ( z,α − n ) ( ζ ) and A n ( ζ ) = X z ∈ A n χ B Z ( z,α − n ) ( ζ )are ν -measurable, and hence so is u n . We shall show that for ν -a.e. ζ ∈ Z , the limit˜ u ( ζ ) = lim n →∞ u n ( ζ ) (11.5)exists and defines the trace ˜ u : Z → R . To this end, note that ( z, j ) ∼ ( y, j + 1)whenever z ∈ A j ( ζ ) and y ∈ A j +1 ( ζ ), j = 0 , , ... , since ζ ∈ B ( z, α − j ) ∩ B ( y, α − j − ).Also, 1 A j ( ζ ) A j +1 ( ζ ) ≤ . We then have for each j , | u j ( ζ ) − u j +1 ( ζ ) | ≤ X z ∈ A j ( ζ ) X y ∈ A j +1 ( ζ ) | u (( z, j )) − u (( y, j + 1)) |≤ X z ∈ A j ( ζ ) X y ∈ A j +1 ( ζ ) Z [( z,j ) , ( y,j +1)] g ds ε . (11.6)On the edge E = [( z, j ) , ( y, j + 1)], we have by (10.7) that ds ε ≃ e − εj d L = α − j dµ β µ β ( E ) . (11.7)If p > 1, then (11.7) and H¨older’s inequality applied to (11.6) give | u j ( ζ ) − u j +1 ( ζ ) | ≤ α − j X z ∈ A j ( ζ ) X y ∈ A j +1 ( ζ ) Z [( z,j ) , ( y,j +1)] g dµ β ≤ α − j X z ∈ A j ( ζ ) X E ∈E ( z,j ) (cid:18)Z E g p dµ β (cid:19) /p , (11.8) xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 39 where E ( z, j ) consists of all downward-directed edges emanating from the vertex( z, j ). Choose 0 < κ < θp and insert α − jκ/p α jκ/p into (11.8). Summing over j ,together with another use of H¨older’s inequality, this time on the sum, shows thatfor all m > n ≥ | u n ( ζ ) − u m ( ζ ) | ≤ m − X j = n | u j ( ζ ) − u j +1 ( ζ ) | (11.9) ≤ ∞ X j = n α − jκ/p α − j (1 − κ/p ) X z ∈ A j ( ζ ) X E ∈E ( z,j ) (cid:18)Z E g p dµ β (cid:19) /p . α − nκ/p (cid:18) ∞ X j = n α − j ( p − κ ) X z ∈ A j ( ζ ) X E ∈E ( z,j ) Z E g p dµ β (cid:19) /p , where we have also used the fact that (cid:18) ∞ X j = n α − jκ/ ( p − (cid:19) − /p ≃ α − nκ/p , together with A j ( ζ ) ≤ K and E ( z, j ) ≤ K . For p = 1 the estimate is simplerand H¨older’s inequality is not needed, and the above estimate holds as well. Weshall now see that (11.9) tends to zero as m > n → ∞ for ν -a.e. ζ ∈ Z . Thus, thesequence { u n ( ζ ) } ∞ n =0 is a Cauchy sequence, and has a limit as n → ∞ , for ν -a.e. ζ ∈ Z .To this end, note that for E ∈ E ( z, j ), Z E g p dµ β ≃ α jβ/ε ν ( B Z ( z, α − j )) Z X V g ( x ) p χ E ( x ) dµ β ( x ) , (11.10)where X V denotes the union of all vertical edges in X . Also, z ∈ A j ( ζ ) if and onlyif z ∈ A j and ζ ∈ B Z ( z, α − j ). Integrating over all ζ ∈ Z we then obtain from (11.9)by means of Tonelli’s theorem that Z Z | u m ( ζ ) − u n ( ζ ) | p dν ( ζ ) . α − nκ Z Z ∞ X j = n X z ∈ A j α − j ( p − β/ε − κ ) ν ( B Z ( z, α − j )) χ B Z ( z,α − j ) ( ζ ) × X E ∈E ( z,j ) Z X V g ( x ) p χ E ( x ) dµ β ( x ) dν ( ζ )= α − nκ Z X V g ( x ) p ∞ X j = n α − j ( p − β/ε − κ ) × X z ∈ A j X E ∈E ( z,j ) Z Z χ B Z ( z,α − j ) ( ζ ) ν ( B Z ( z, α − j )) dν ( ζ ) χ E ( x ) dµ β ( x ) . (11.11)The integral over Z is clearly equal to 1. Moreover, for a.e. x ∈ X , χ E ( x ) = 0 only if x ∈ E ∈ E ( z, j ) with j = π ( x ) , and so for a.e. x ∈ X we have X z ∈ A j X E ∈E ( z,j ) Z Z χ B Z ( z,α − j ) ( ζ ) ν ( B Z ( z, α − j )) dν ( ζ ) χ E ( x ) = X z ∈ A j X E ∈E ( z,j ) χ E ( x )= χ { y ∈ X V : π ( y )= j } ( x ) . (11.12) We therefore conclude that Z Z | u m ( ζ ) − u n ( ζ ) | p dν ( ζ ) . α − nκ Z { y ∈ X V : π ( y ) ≥ n } g ( x ) p α − π ( x )( p − β/ε − κ ) dµ β ( x ) . Since p − β/ε − κ ≥ θp − κ > 0, we obtain that Z Z | u m ( ζ ) − u n ( ζ ) | p dν ( ζ ) . α − n ( p − β/ε ) Z X V g p dµ β → , as m > n → ∞ . Hence, the sequence { u n } ∞ n =0 is a Cauchy sequence both in L p ( Z ) and ν -a.e. in Z (recall that we have m > n in the above computations). The limit (11.5) thereforeexists for ν -a.e. ζ ∈ Z , and ˜ u ∈ L p ( Z ).This also shows, by letting n = 0 and u m → ˜ u , that (cid:18)Z Z | ˜ u − u ( v ) | p dν (cid:19) /p . Z X V g p dµ β , where v = ( z , k ˜ u k L p ( Z ) . | u ( v ) | + k g k L p ( X ε ,µ β ) holds in (11.3). For the second inequality, recall the notion of capacity from Defini-tion 9.3. Since | u ( v ) | p C X ε p ( { v } ) ≤ k u k pN ,p ( X ε ,µ β ) by the definition of C X ε p ( { v } ),and C X ε p ( { v } ) > 0, by Remark 9.5, we conclude that the second inequality in (11.3)holds as well.To estimate k ˜ u k θ,p , we let m → ∞ in (11.9) to obtain for ν -a.e. ζ ∈ Z and any n ≥ | ˜ u ( ζ ) − u n ( ζ ) | p . α − nκ ∞ X j = n α − j ( p − κ ) X z ∈ A j ( ζ ) X E ∈E ( z,j ) Z E g p dµ β . (11.13)A similar estimate holds for ν -a.e. ξ ∈ Z . As in Lemma 4.3, we let l ≥ α − l ≤ τ − 1. Also let Z n ( ζ ) = { ξ ∈ Z : α − n − l − < d Z ( ξ, ζ ) ≤ α − n − l } , n = 1 , , ... , and Z ( ζ ) = Z \ B Z ( ζ, α − l − ). Note that ξ ∈ Z n ( ζ ) if and only if ζ ∈ Z n ( ξ ), inwhich case also ζ ∈ τ B Z ( z, α − n ) ∩ τ B Z ( y, α − n ) and thus ( z, n ) ∼ ( y, n ) for all z ∈ A n ( ζ ) and y ∈ A n ( ξ ) with y = z . Hence for all ξ ∈ Z n ( ζ ), n = 1 , ... , | u n ( ζ ) − u n ( ξ ) | p . X z ∈ A n ( ζ ) X y ∈ A n ( ξ ) y = z | u (( z, n )) − u (( y, n )) | p , (11.14)while u ( ζ ) = u ( v ) = u ( ξ ) for all ζ, ξ ∈ Z . H¨older’s inequality and (11.7) with E = [( z, n ) , ( y, n )] give | u (( z, n )) − u (( y, n )) | p ≤ (cid:18)Z E g ds ε (cid:19) p . α − np Z E g p dµ β . (11.15)Next, note that | ˜ u ( ζ ) − ˜ u ( ξ ) | p . | ˜ u ( ζ ) − u n ( ζ ) | p + | u n ( ζ ) − u n ( ξ ) | p + | u n ( ξ ) − ˜ u ( ξ ) | p , (11.16)and that each of the three terms can be estimated with the aid of (11.13)–(11.15).We shall insert (11.16) into the Besov norm k ˜ u k pθ,p = Z Z Z Z \{ ζ } | ˜ u ( ξ ) − ˜ u ( ζ ) | p d Z ( ξ, ζ ) θp dν ( ξ ) dν ( ζ ) ν ( B Z ( ζ, d Z ( ξ, ζ ))) , xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 41 and obtain three terms corresponding to the three terms on the right-hand sideof (11.16). We next use the comparisons d Z ( ξ, ζ ) θp ≃ α − nθp and ν ( B Z ( ζ, d Z ( ξ, ζ ))) ≃ ν ( B Z ( ξ, d Z ( ξ, ζ ))) ≃ ν ( B Z ( ζ, α − n )) ≃ ν ( B Z ( ξ, α − n ))whenever ξ ∈ Z n ( ζ ) (or equivalently, ζ ∈ Z n ( ξ )). We then get k ˜ u k pθ,p . I + II + III , where I := Z Z ∞ X n =0 Z Z n ( ζ ) | ˜ u ( ζ ) − u n ( ζ ) | p α − nθp dν ( ξ ) dν ( ζ ) ν ( B Z ( ζ, α − n )) ,II := Z Z ∞ X n =0 Z Z n ( ζ ) | u n ( ζ ) − u n ( ξ ) | p α − nθp dν ( ξ ) dν ( ζ ) ν ( B Z ( ζ, α − n )) ,III := Z Z ∞ X n =0 Z Z n ( ξ ) | ˜ u ( ξ ) − u n ( ξ ) | p α − nθp dν ( ζ ) dν ( ξ ) ν ( B Z ( ξ, α − n )) . Observe that III is the same as I once the roles of ζ and ξ are switched, and soit suffices to find estimates for I and II . Using (11.13)–(11.15), we find that I . Z Z ∞ X n =0 α − n ( κ − θp ) ν ( B Z ( ζ, α − n )) Z Z n ( ζ ) ∞ X j = n α − j ( p − κ ) × X z ∈ A j ( ζ ) X E ∈E ( z,j ) Z E g ( x ) p dµ β ( x ) dν ( ξ ) dν ( ζ ) =: I,II . Z Z ∞ X n =1 α − n ( p − θp ) ν ( B Z ( ζ, α − n )) × Z Z n ( ζ ) X z ∈ A n ( ζ ) X y ∈ A n ( ξ ) y = z Z [( z,n ) , ( y,n )] g ( x ) p dµ β ( x ) dν ( ξ ) dν ( ζ ) =: II. To estimate I , we use (11.10) and that z ∈ A j ( ζ ) if and only if z ∈ A j and ζ ∈ B Z ( z, α − j ). Now an argument using Tonelli’s theorem as in the verificationof (11.11) yields that I ≃ ∞ X n =0 α − n ( κ − θp ) ∞ X j = n α − j ( p − β/ε − κ ) X z ∈ A j X E ∈E ( z,j ) × Z Z χ B Z ( z,α − j ) ( ζ ) ν ( B Z ( z, α − j )) Z Z n ( ζ ) dν ( ξ ) ν ( B Z ( ζ, α − n )) dν ( ζ ) Z X V g ( x ) p χ E ( x ) dµ β ( x ) . Since Z n ( ζ ) ⊂ B Z ( ζ, α − n ), the integral over Z n ( ζ ) followed by the integral over Z is clearly at most 1. Another use of Tonelli’s theorem therefore implies that I . Z X V g ( x ) p ∞ X n =0 α − n ( κ − θp ) ∞ X j = n α − j ( p − β/ε − κ ) X z ∈ A j X E ∈E ( z,j ) χ E ( x ) dµ β ( x ) . The last three sums are simplified using the last identity in (11.12) and we obtain I . Z X V g ( x ) p α − π ( x )( p − β/ε − κ ) ∞ X n =0 α − n ( κ − θp ) dµ β ( x ) ≃ Z X V g ( x ) p α − π ( x )( p − β/ε − θp ) dµ β ( x ) , because of the choices κ < θp and α > 1. Since p − β/ε − θp ≥ 0, this yields I . Z X V g p dµ β . To estimate II , we proceed similarly. As in (11.10), we have that when ( z, n ) ∼ ( y, n ), Z [( z,n ) , ( y,n )] g p dµ β ≃ α nβ/ε ν ( B Z ( z, α − n )) Z X H g ( x ) p χ [( z,n ) , ( y,n )] ( x ) dµ β ( x ) , where X H denotes the union of all horizontal edges in X .Moreover, z ∈ A n ( ζ ) and y ∈ A n ( ξ ) if and only if z, y ∈ A n , ζ ∈ B Z ( z, α − n )and ξ ∈ B Z ( y, α − n ). Tonelli’s theorem then yields that II ≃ ∞ X n =1 α − n ( p − β/ε − θp ) X z,y ∈ A n ( z,n ) ∼ ( y,n ) Z Z χ B Z ( z,α − n ) ( ζ ) ν ( B Z ( z, α − n )) Z Z n ( ζ ) χ B Z ( y,α − n ) ( ξ ) ν ( B Z ( ζ, α − n )) dν ( ξ ) dν ( ζ ) × Z X H g ( x ) p χ [( z,n ) , ( y,n )] ( x ) dµ β ( x ) . Since Z n ( ζ ) ⊂ B Z ( ζ, α − n ), the integral over Z n ( ζ ) followed by the integral over Z is at most 1, and another use of Tonelli’s theorem shows that II . Z X H g ( x ) p ∞ X n =1 α − n ( p − β/ε − θp ) X z,y ∈ A n ( z,n ) ∼ ( y,n ) χ [( z,n ) , ( y,n )] ( x ) dµ β ( x ) . Moreover, χ [( z,n ) , ( y,n )] ( x ) = 0 if only if x ∈ [( z, n ) , ( y, n )], in which case also n = π ( x ). We therefore conclude that II . Z X H g ( x ) p α − π ( x )( p − β/ε − θp ) dµ β ( x ) ≤ Z X H g p dµ β , because p − β/ε − θp ≥ 0. Combining the estimates for I and II gives the desiredbound (11.2).The fact that ˜ u = ˆ u | Z when u is Lipschitz continuous on X ε follows from thedefinition of ˜ u and the fact that u has a unique Lipschitz extension to X ε .Recall the notion of capacity from Definition 9.3. The following propositionshows that the boundary measure ν on Z = ∂ ε X is absolutely continuous withrespect to the C X ε p -capacity. Note that points in X have positive C X ε p -capacity,but that it is possible to have nonempty subsets of ∂ ε X = Z with zero capacity. Proposition 11.2. Let E ⊂ ∂ ε X . If p > β/ε and C X ε p ( E ) = 0 , then ν ( E ) = 0 .Proof. Since C X ε p is an outer capacity by Theorem 9.6, there are open sets G j ⊃ E such that C X ε p ( G j ) < /j . Then E ′ := ∞ \ j =1 G j ⊃ E is a Borel set with zero capacity. Let K ⊂ E ′ be compact. Because µ β is doublingand supports a p -Poincar´e inequality on X ε , it follows from Kallunki–Shanmuga-lingam [31, Theorem 1.1] (or [2, Theorem 6.7 (xi)]) that there are u k ∈ Lip( X ε ) xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 43 such that u k = 1 on K and k u k k N ,p ( X ε ,µ β ) < /k , k = 1 , , ... . By the last part ofTheorem 11.1 with θ = 1 − β/εp > ν ( K ) /p ≤ lim k →∞ k u k k L p ( Z ) . lim k →∞ k u k k N ,p ( X ε ,µ β ) = 0 , i.e., ν ( K ) = 0. Since E ′ is a Borel set and ν is a Borel regular measure, we concludethat ν ( E ) ≤ ν ( E ′ ) = sup K ⊂ E ′ compact ν ( K ) = 0 . The following result is a refinement of Theorem 11.1. In the case of regulartrees, it provides a more precise trace result than Proposition 6.1 in Bj¨orn–Bj¨orn–Gill–Shanmugalingam [3]. Recall that by Theorem 7.1, every rooted tree X can beseen as a hyperbolic filling of its uniformized boundary ∂ ε X . Theorem 11.3. Let u ∈ e N ,p ( X ε , µ β ) and < θ ≤ − β/εp . Then u has anextension ˆ u ∈ e N ,p ( X ε , µ β ) . Furthermore, the restriction ˜ u := ˆ u | Z agrees with thetrace of u defined earlier ν -a.e. in Z , and belongs to B θp,p ( Z ) with the ( semi ) normestimates k ˜ u k θ,p . k g u k L p ( X ε ,µ β ) and k ˜ u k L p ( Z ) . | u ( x ) | + k g u k L p ( X ε ,µ β ) . k u k N ,p ( X ε ,µ β ) . Moreover, for C X ε p -q.e. ( and thus ν -a.e. ) ζ ∈ Z we have that lim r → + Z X ε ∩ B ε ( ζ,r ) | u − ˜ u ( ζ ) | p dµ β = 0 . (11.17)Note that the extension ˆ u is not unique, but it is unique up to sets of capacityzero and thus ν -a.e. (by Proposition 11.2), since if ˆ u and ˆ u are two extensions,then they are equal µ β -a.e., and thus C X ε p -q.e. We may therefore take the restrictionof any such extension. The key observation that makes the above statement trueis that the representatives in Newtonian spaces are equal q.e., not just a.e. as forstandard Sobolev spaces.The last claim of the above theorem tells us that the trace of a function in N ,p ( X ε , µ β ), as constructed in Theorem 11.1, agrees with other notions of tracesin the current literature, see e.g. Mal´y [39]. Proof. By Theorems 5.1 and 10.3, X ε is a uniform domain in X ε and µ β is doublingand supports a p -Poincar´e inequality on X ε . Thus by Proposition 5.9 in Bj¨orn–Shanmugalingam [7], X ε is an extension domain, and thus u has an extension to X ε , denoted ˆ u ∈ N ,p ( X ε , µ β ).By Shanmugalingam [45, Theorem 4.1 and Corollary 3.9], there is a sequence u j ∈ Lip( X ε ) such that k u j − ˆ u k N ,p ( X ε ,µ β ) → u j ( x ) → ˆ u ( x ) for C X ε p -q.e. x ∈ X ε , as j → ∞ . Let ˜ u := ˆ u | Z and ˜ u j := u j | Z . By Proposition 11.2, ˜ u j ( ζ ) → ˜ u ( ζ )for ν -a.e. ζ ∈ Z . Moreover, by Theorem 11.1, { ˜ u j } ∞ j =1 is a Cauchy sequence in thenorm k · k B θp,p ( Z ) . By Remark 9.8, B θp,p ( Z ) is complete with respect to k · k B θp,p ( Z ) ,and hence we see that k ˜ u j − ˜ u k B θp,p ( Z ) → j → ∞ , with the (semi)norm estimatesfrom Theorem 11.1 preserved for ˜ u .By [2, Theorem 5.62] or [27, Theorem 9.2.8] (for p > 1) and Kinnunen–Korte–Shanmugalingam–Tuominen [32, Theorem 4.1 and Remark 4.7] (for p = 1, seebelow) we know that C X ε p -q.e. point in X ε is an L p ( µ β )-Lebesgue point of ˆ u ; hence(11.17) holds because µ β ( ∂ ε X ) = 0.In [32, p. 404] it is assumed that µ ( X ) = ∞ , which is used in their proof ofthe boxing inequality. In M¨ak¨al¨ainen [38], the boxing inequality is proved alsowhen µ ( X ) < ∞ , and thus the Lebesgue point result in [32] holds also here where µ β ( X ε ) < ∞ . 12. Extension from Z to its hyperbolic filling X Recall the standing assumptions from Sections and . Theorem 11.3 related the Newtonian space N ,p ( X ε , µ β ) to a certain range ofBesov spaces of functions on Z ≡ ∂ ε X . The principal goal of this section is to finda counterpart of this theorem in the opposite direction. This is the purpose of thetheorem below. Theorem 12.1. For f ∈ B θp,p ( Z ) , consider the extension Ef (( z, n )) := Z B Z ( z,α − n ) f dν, if ( z, n ) ∈ V ⊂ X, extended piecewise linearly ( with respect to d ε ) to each edge in X ε , and then to theboundary ∂ ε X by letting Ef ( ζ ) = lim sup r → + Z B ε ( ζ,r ) Ef dµ β , ζ ∈ ∂ ε X, (12.1) so that Ef : X ε → [ −∞ , ∞ ] .If θ ≥ − β/pε , then Ef ∈ N ,p ( X ε , µ β ) with Z X ε g pEf dµ β . k f k pθ,p and Z X ε | Ef | p dµ β . Z Z | f | p dν. (12.2) Moreover, if ζ ∈ Z is an L q ( ν ) -Lebesgue point of f for some q ≥ then ζ is an L q ( µ β ) -Lebesgue point of Ef , and Ef ( z ) = f ( z ) . Let ζ be an L ( ν ) -Lebesgue point of f . Then for any choice of z n ∈ A n with d Z ( z n , ζ ) < α − n for n = 1 , , ... , we have, lim n →∞ Ef (( z n , n )) = f ( ζ ) . Note that E is a linear operator. Proof. If v = ( z, n ) ∼ ( y, m ) = w , then | m − n | ≤ ε = log α , d ε ( v, w ) ≃ e − εn = α − n . The function given for x ∈ [ v, w ] by g [ v,w ] ( x ) := | Ef ( v ) − Ef ( w ) | d ε ( v, w )is an upper gradient of Ef on [ v, w ] with respect to the uniformized metric d ε . Notethat g [ v,w ] is a constant function. Because of | m − n | ≤ z, n ) ∼ ( y, m ), wehave for all η ∈ B Z ( z, α − n ) that B Z ( y, α − m ) ⊂ τ B Z ( η, α − n ) . Hence, g [ v,w ] ≃ α n (cid:12)(cid:12)(cid:12)(cid:12)Z B Z ( z,α − n ) f ( ζ ) dν ( ζ ) − Z B Z ( y,α − m ) f ( η ) dν ( η ) (cid:12)(cid:12)(cid:12)(cid:12) . α n Z B Z ( z,α − n ) Z B Z ( y,α − m ) | f ( ζ ) − f ( η ) | dν ( η ) dν ( ζ ) . α n Z B Z ( z, τα − n ) Z B Z ( η, τα − n ) | f ( ζ ) − f ( η ) | dν ( η ) dν ( ζ ) . xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 45 Now by H¨older’s inequality, we see that g p [ v,w ] . α np (1 − θ ) Z B Z ( z, τα − n ) Z B Z ( η, τα − n ) | f ( ζ ) − f ( η ) | p α − npθ dν ( η ) dν ( ζ ) . Therefore, letting g : X → R be given by g = g [ v,w ] on each edge [ v, w ], and notingfrom (10.6)–(10.7) that dµ β ≃ α − βn/ε ν ( B Z ( z, α − n )) d L on [ v, w ] with n = π ( v ),we obtain Z [ v,w ] g p dµ β . α n ( p (1 − θ ) − β/ε ) Z B Z ( z, τα − n ) Z B Z ( η, τα − n ) | f ( ζ ) − f ( η ) | p α − npθ dν ( η ) dν ( ζ ) , where z = π ( v ) ∈ A n ⊂ Z . For each nonnegative integer n set X ( n ) := { x ∈ X : n ≤ π ( x ) < n + 1 } , where π ( x ) is as in (11.1). By Proposition 4.5, each vertex in X has degree at most K and thus we get integrating over X ( n ) that Z X ( n ) g p dµ β ≤ X z ∈ A n X V ∋ w ∼ ( z,n ) Z [( z,n ) ,w ] g p dµ β . α n ( p (1 − θ ) − β/ε ) X z ∈ A n Z B Z ( z, τα − n ) Z B Z ( η, τα − n ) | f ( ζ ) − f ( η ) | p α − npθ dν ( η ) dν ( ζ ) . α n ( p (1 − θ ) − β/ε ) Z Z Z B Z ( η, τα − n ) | f ( ζ ) − f ( η ) | p α − npθ dν ( η ) dν ( ζ ) . In the last line we used the fact that the balls B Z ( z, τ α − n ), z ∈ A n , have abounded overlap in Z because of the doubling property of Z . As X = S ∞ n =0 X ( n )with X ( n ) ∩ X ( m ) = ∅ if m = n , it follows that Z X ε g p dµ β . ∞ X n =0 α n ( p (1 − θ ) − β/ε ) Z Z Z B Z ( η, τα − n ) | f ( ζ ) − f ( η ) | p α − npθ dν ( η ) dν ( ζ ) . If θ ≥ − β/pε , it then follows from Lemma 9.9 that Z X ε g p dµ β . k f k pθ,p < ∞ . (12.3)As µ β supports a 1-Poincar´e inequality on X ε , by Theorem 10.3, and X ε is bounded,it follows that Ef ∈ N ,p ( X ε , µ β ).As in the proof of Theorem 11.3, we have an extension u ∈ N ,p ( X ε , µ β ) of Ef ,and C X ε p -q.e. point in X ε is a Lebesgue point of u . As µ β ( ∂ ε X ) = 0, we see that u ( x ) = Ef ( x ) for C X ε p -q.e. x ∈ ∂ ε X , where Ef | ∂ ε X is given by (12.1). Hence Ef is also in N ,p ( X ε , µ β ). Since X ε is open in X ε , we see that the minimal p -weakupper gradients of Ef with respect to X ε and X ε coincide almost everywhere in X ε , and thus the first inequality in (12.2) follows from (12.3).To control the L p -norm of Ef as stated in the theorem, note that for v = ( z, n ) ∼ w = ( y, m ), Z [ v,w ] | Ef | p dµ β ≤ µ β ([ v, w ])[ | Ef ( v ) | p + | Ef ( w ) | p ] ≤ µ β ([ v, w ]) (cid:18)Z B Z ( z,α − n ) | f | p dν + Z B Z ( y,α − m ) | f | p dν (cid:19) . µ β ([ v, w ]) Z B Z ( z, τα − n ) | f | p dν. Therefore, for each nonnegative integer n , with X ( n ) as above, we have that Z X ( n ) | Ef | p dµ β . α − nβ/ε X z ∈ A n ν ( B Z ( z, α − n )) Z B Z ( z, τα − n ) | f | p dν . α − nβ/ε Z Z | f | p dν. It follows that Z X ε | Ef | p dµ β . k f k pL p ( Z ) ∞ X n =0 α − nβ/ε . k f k pL p ( Z ) as desired.Assume that q ≥ ζ ∈ Z is an L q ( ν )-Lebesgue point of f . Let N ≥ x ∈ X such that d ε ( x, ζ ) < r := α − N /ε .If x belongs to an edge [ v, w ], then at least one of the vertices also belongs to B ε ( ζ, r ),say w = ( y, m ), and hence α − m = e − εm = εd ε ( w ) ≤ εd ε ( w, ζ ) < εr = α − N , from which it follows that m ≥ N + 1 and thus n ≥ N , where v = ( z, n ). Inparticular, d ε ( z, v ) = d ε ( v ) = e − εn /ε ≤ r and d ε ( v, x ) ≤ e − εN = εr . Proposition 4.4then yields d Z ( z, ζ ) ≤ τ αd ε ( z, ζ ) ≤ τ α ( d ε ( z, v ) + d ε ( v, x ) + d ε ( x, ζ )) < τ α (2 + ε ) r, and similarly d Z ( y, ζ ) < τ α (2 + ε ) r . Since Ef ( x ) is a convex combination of Ef ( v )and Ef ( w ), we have Z [ v,w ] | Ef − f ( ζ ) | q dµ β ≤ ( | Ef ( v ) − f ( ζ ) | q + | Ef ( w ) − f ( ζ ) | q ) µ β ([ v, w ]) , where by the definition of Ef and H¨older’s inequality, | Ef ( v ) − f ( ζ ) | q = (cid:12)(cid:12)(cid:12)(cid:12)Z B Z ( z,α − n ) f dν − f ( ζ ) (cid:12)(cid:12)(cid:12)(cid:12) q ≤ Z B Z ( z,α − n ) | f − f ( ζ ) | q dν. Noting that µ β ([ v, w ]) ≃ ρ β ( v )ˆ µ ( v ) = e − βn ν ( B Z ( z, α − n ))and that every vertex belongs to at most a bounded number of edges (by Proposi-tion 4.5), we conclude that Z B ε ( ζ,r ) | Ef − f ( ζ ) | q dµ β . X n ≥ N e − βn X z ∈ A n ∩ B Z ( ζ, τα (2+ ε ) r ) Z B Z ( z,α − n ) | f − f ( ζ ) | q dν. Since for each n ≥ N we have that α − n ≤ α − N = εr and the balls B Z ( z, α − n ), z ∈ A n , have bounded overlap in Z , we obtain Z B ε ( ζ,r ) | Ef − f ( ζ ) | q dµ β . X n ≥ N e − βn Z B Z ( ζ, τα (1+ ε ) r ) | f − f ( ζ ) | q dν ≃ e − βN Z B Z ( ζ, τα (1+ ε ) r ) | f − f ( ζ ) | q dν, where e − βN = ( α − N ) β/ε = ( εr ) β/ε . Dividing by µ β ( B ε ( ζ, r )) ≃ ( εr ) β/ε ν ( B Z ( ζ, εr ))(because of Corollary 10.5) and letting N → ∞ shows that ζ is an L q ( µ β )-Lebesgue xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 47 point of Ef . That Ef ( ζ ) = f ( ζ ) now follows directly from the definition of Ef ( ζ )in (12.1) and by considering the case q = p in the above discussion (recall that f isnecessarily in L p ( ν ) and so ν -a.e. point in Z is an L p ( ν )-Lebesgue point of f ).Moreover, with z n as in the final claim of the theorem for n = 1 , , ... , thedoubling property of ν and the fact that B Z ( z n , α − n ) ⊂ B Z ( ζ, α − n ) yield | Ef (( z n , n )) − f ( ζ ) | = (cid:12)(cid:12)(cid:12)(cid:12)Z B Z ( z n ,α − n ) [ f ( y ) − f ( ζ )] dν ( y ) (cid:12)(cid:12)(cid:12)(cid:12) . Z B Z ( ζ, α − n ) | f ( y ) − f ( ζ ) | dν ( y ) → , as n → ∞ . 13. Properties of Besov functions on Z Recall the standing assumptions from Sections and . Since ε = log α , fromProposition we know that ∂ ε X can be identified with Z in a biLipschitz fashion.In this section we also fix < θ < and let β = εp (1 − θ ) . In this section we will only consider the Besov spaces on Z that arise as tracesof Newtonian functions on X ε as in Theorem 1.1. This makes it possible to derivevarious regularity properties for B θp,p ( Z ) from from the theory of Newtonian spaces. Proposition 13.1. If < θ < then Lipschitz functions are dense in B θp,p ( Z ) .Proof. Equip the uniformized hyperbolic filling X ε with the measure µ β , where β = εp (1 − θ ). Theorems 11.3 and 12.1 tell us that B θp,p ( Z ) is the trace space of N ,p ( X ε , µ β ), with comparable norms.Since µ β is doubling and supports a 1-Poincar´e inequality on X ε , it follows fromShanmugalingam [45, Theorem 4.1 and Corollary 3.9] that Lipschitz functions aredense in N ,p ( X ε , µ β ). Their restrictions to Z are then dense in B θp,p ( Z ). Proposition 13.2. Let E ⊂ Z . Then Cap B θp,p ( Z ) ( E ) ≃ C X ε p ( E ) .Proof. Let u ∈ B θp,p ( Z ) be admissible in the definition of Cap B θp,p ( Z ) ( E ), i.e. u ≥ ν -a.e. in an open neighborhood G ⊂ Z of E . By truncation and redefinition on aset of ν -measure zero, we may assume that u ≡ G . Let Eu ∈ N ,p ( X ε , µ β )be its extension as guaranteed by Theorem 12.1. As all points in G are Lebesguepoints for u , we see that Eu ≡ u ≡ G . Hence Eu is admissible in computing C X ε p ( E ), and so by Theorem 12.1, C X ε p ( E ) ≤ k Eu k pN ,p ( X ε ,µ β ) . k u k pB θp,p ( Z ) . Taking infimum over all u admissible in the definition of Cap B θp,p ( Z ) ( E ) proves oneinequality in the statement of the lemma.Conversely, since C X ε p is an outer capacity by Theorems 9.6 and 10.3, for each η > U ⊂ X ε with E ⊂ U and a function u ∈ e N ,p ( X ε , µ β )such that u ≥ U and k u k N ,p ( X ε ,µ β ) < C X ε p ( E ) + η. By Theorem 11.3, the function f = u | Z ∈ B θp,p ( Z ) with k f k pL p ( Z ) + k f k pθ,p . k u k pN ,p ( X ε ,µ β ) < C X ε p ( E ) + η. As f ≥ G := U ∩ Z , letting η → We have now shown that for subsets of Z the two capacities are comparable.Next we turn our attention to the matter of continuity properties of Besov functions.The following result shows that functions in B θp,p ( Z ) have representatives that are quasicontinuous with respect to the Besov capacity, i.e. such that for each η > G ⊂ Z with Cap B θp,p ( Z ) ( G ) < η such that f | Z \ G is continuous. Proposition 13.3. Let f ∈ B θp,p ( Z ) . Then there is a Cap B θp,p ( Z ) -quasicontinuousfunction f ∈ B θp,p ( Z ) such that f = f ν -a.e. in Z .Proof. Given such a function f , let Ef ∈ N ,p ( X ε , µ β ) be its extension givenby Theorem 12.1. Then f := Ef | Z = f ν -a.e. By Theorems 9.6 and 10.3, Ef is C X ε p -quasicontinuous, i.e. for each η > U ⊂ X ε with C X ε p ( U ) < η such that Ef | X ε \ U is continuous. By choosing G = U ∩ Z , we getthat f | Z \ G is continuous. Moreover, by Proposition 13.2, we see thatCap B θp,p ( Z ) ( G ) ≃ C X ε p ( G ) ≤ C X ε p ( U ) < η, which completes the proof.The following result shows that Besov functions have Lebesgue points q.e., pro-vided that the measure on Z satisfies a reverse-doubling property. Proposition 13.4. Assume that ν satisfies (10.13) with s ν > and that there issome η > such that ν ( B Z ( ζ, r ′ )) ν ( B Z ( ζ, r )) . (cid:16) r ′ r (cid:17) η (13.1) for all ζ ∈ Z and all < r ′ ≤ r ≤ Z . Let ≤ q ≤ s ν p/ ( s ν − pθ ) and ˜ u ∈ B θp,p ( Z ) . Then there is a function u ∈ L q ( Z ) such that u = ˜ u ν -a.e. and lim r → + Z B ε ( ζ,r ) ∩ Z | u − u ( ζ ) | q dν = 0 for Cap B θp,p ( Z ) -q.e. ζ ∈ Z . Since ν is doubling, condition (13.1) is equivalent to Z being uniformly perfect,see Mart´ın–Ortiz [41, Lemma 7]. See [47] for a weaker Lebesgue point result when Z is not necessarily uniformly perfect. Embeddings of Besov spaces into L q spaceswere also obtained in Mal´y [39, Corollary 3.18 (i)] via embeddings into Haj lasz–Sobolev spaces. Proposition 13.4 will follow from our trace and extension resultsand the following two-weighted Poincar´e type inequality, which is a special case ofBj¨orn–Ka lamajska [6, Theorem 3.1]. Proposition 13.5. Let ν and µ β be doubling measures on Z = ∂ ε X and X ε , respec-tively. Assume moreover that µ β supports a p -Poincar´e inequality on X ε with dila-tion λ and that ν satisfies the reverse-doubling condition (13.1) . Let ≤ p < q < ∞ and u ∈ N ,p ( X ε , µ β ) be such that ν -a.e. z ∈ Z is a µ β -Lebesgue point of u . Thenfor all balls B = B ε ( ζ, r ) with ζ ∈ Z , (cid:18)Z B ∩ Z | u − u B,µ β | q dν (cid:19) /q . Θ q ( r ) (cid:18)Z λB g pu dµ β (cid:19) /p , where Θ q ( r ) := sup <ρ ≤ r sup z ∈ B ∩ Z ρν ( B ε ( z, ρ )) /q µ β ( B ε ( z, ρ )) /p . xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 49 Proof of Proposition . By H¨older’s inequality, it suffices to consider the case of q > p . Using Theorem 12.1, we can find a function u ∈ N ,p ( X ε , µ β ) such that u = ˜ u ν -a.e. on Z . By [27, Lemma 9.2.4] and Theorem 11.3, we know that for C X ε p -q.e. ζ ∈ Z ,lim r → + r p Z B ε ( ζ,r ) g pu dµ β = 0 and lim r → + Z B ε ( ζ,r ) u dµ β = u ( ζ ) . (13.2)In particular, Proposition 11.2 shows that ν -a.e. z ∈ Z is a µ β -Lebesgue point of u .Proposition 13.2 shows that (13.2) holds for Cap B θp,p ( Z ) -q.e. ζ ∈ Z . For such ζ ,we have by the Minkowski inequality and Proposition 13.5 that (cid:18)Z B ε ( ζ,r ) ∩ Z | u − u ( ζ ) | q dν (cid:19) /q . (cid:12)(cid:12)(cid:12)(cid:12)Z B ε ( ζ,r ) u dµ β − u ( ζ ) (cid:12)(cid:12)(cid:12)(cid:12) (13.3)+ Θ q ( r ) µ β ( B ε ( ζ, λr )) /p ν ( B ε ( ζ, r )) /q (cid:18)Z B ε ( ζ, λr ) g pu dµ β (cid:19) /p . In view of (13.2) and the definition of Θ q ( r ), it suffices to show that for all 0 < ρ ≤ r ≤ ε X ε and z ∈ B ε ( ζ, r )) ∩ Z , ρν ( B ε ( z, ρ )) /q µ β ( B ε ( z, ρ )) /p µ β ( B ε ( ζ, λr )) /p ν ( B ε ( ζ, r )) /q . r (13.4)with a comparison constant independent of z , ρ and r . Theorem 10.3 and thedoubling property show that ρν ( B ε ( z, ρ )) /q µ β ( B ε ( z, ρ )) /p ≃ ρ − β/εp ν ( B ε ( z, ρ )) /q − /p and µ β ( B ε ( ζ, λr )) /p ν ( B ε ( ζ, r )) /q ≃ r β/εp ν ( B ε ( ζ, r )) /p − /q ≃ r β/εp ν ( B ε ( z, r )) /p − /q . Since 1 /q − /p < ν satisfies (10.13), the required estimate (13.4) holdsbecause 1 − βεp + s ν (cid:18) q − p (cid:19) ≥ . The conclusion u ∈ L q ( Z ) follows by applying (13.3) to r = 2 diam X ε .Even though we have so far only considered compact Z , we can now apply Propo-sition 13.4 to obtain the following improvement of Netrusov’s result [42, Proposi-tion 1.4] in R n , which was obtained for q < np/ ( n − pθ ). The lift from the compactto the unbounded case is somewhat subtle since the Besov norm is nonlocal. Proposition 13.6. Assume that n ≥ and that pθ < n . Let q = np/ ( n − pθ ) , ˜ u ∈ B θp,p ( R n ) , u ( ζ ) := lim sup r → + ˜ u B ( ζ,r ) , and E = (cid:26) ζ : lim sup r → + Z | u ( x ) − u ( ζ ) | q dx > (cid:27) be the set of non- L q -Lebesgue points for u . Then Cap B θp,p ( R n ) ( E ) = 0 .Proof. For r > 0, let Z r = B (0 , r ). Observe that for x ∈ Z r and 0 < ρ ≤ r wehave m ( B ( x, ρ ) ∩ Z r ) ≃ m ( B ( x, ρ )) ≃ ρ n . It follows that B θp,p ( R n ) ⊂ B θp,p ( Z r ) for all r > 0. Since Cap B θp,p ( R n ) is comparable to a countably subadditive Besovcapacity on R n , see Adams–Hedberg [1, Propositions 2.3.6 and 4.4.3], it sufficesto show that Cap B θp,p ( R n ) ( E ∩ B (0 , r )) = 0 for all r ≥ 1. Fix ε > R ≥ r . As u ∈ B θp,p ( Z R ), it follows from Proposition 13.4 that Cap B θp,p ( Z R ) ( E ∩ B (0 , r )) = 0. Proposition 13.2, together with [2, Lemma 6.15 and Theorem 4.21]and Theorem 11.3, now implies that there is a function v : R n → [0 , 1] with supp v ⊂ B (0 , r ) such that v ≥ E ∩ B (0 , r ) and k v k pB θp,p ( Z R ) < ε. Now k v k pθ,p + k v k pL p ( R n ) . k v k pB θp,p ( Z R ) + Z R n \ Z R Z B (0 , r ) v ( ζ ) | ζ − ξ | pθ + n dζ dξ . ε + Z R n \ Z R (2 r ) n ( | ξ | − r ) pθ + n dξ → ε, as R → ∞ . Hence Cap B θp,p ( R n ) ( E ∩ B (0 , r )) . ε , and letting ε → Z , and essentiallyrecovers Corollary 3.18 (iii) in Mal´y [39]. Proposition 13.7. Assume that the measure ν on Z satisfies (10.13) for all ζ ∈ Z and < r ′ ≤ r ≤ diam Z , with exponent s ν . Let s β = max { , p (1 − θ ) + s ν } . If p > s β then every f ∈ B θp,p ( Z ) has a ν -a.e. representative which is (1 − s β /p ) -H¨oldercontinuous on Z .Proof. By Theorem 12.1, there is a function u ∈ N ,p ( X ε , µ β ) such that u | Z = fν -a.e. By Lemma 10.6 and by β = εp (1 − θ ), µ β satisfies the dimension condition(10.14).Since p > s β , functions in N ,p ( X ε , µ β ) are (1 − s β /p )-H¨older continuous withrespect to d ε , by [2, Corollary 5.49] or [27, Theorem 9.2.14]. (There is a missinglocal compactness assumption in [27, Theorem 9.2.14].) It follows that the trace u | Z is (1 − s β /p )-H¨older continuous with respect to d Z .Elementary calculations show that Proposition 13.7 applies in the following twocases with 0 < θ < p > s ν /θ : • If p ≥ (1 − s ν ) / (1 − θ ), then s β = p (1 − θ ) + s ν ≥ − s β /p = θ − s ν /p .Hence every f ∈ B θp,p ( Z ) has a ν -a.e. representative which is ( θ − s ν /p )-H¨oldercontinuous. • If 1 < p < (1 − s ν ) / (1 − θ ) (which necessarily implies that s ν < θ ), then p (1 − θ ) + s ν < s β and every f ∈ B θp,p ( Z ) has a ν -a.e. representativewhich is (1 − /p )-H¨older continuous.If θ ≥ B θp,p ( Z ) ⊂ B θ ′ p,p ( Z ) for every 0 < θ ′ < θ replaced by θ ′ imply (upon letting θ ′ → • If p > s ν ≥ 1, then every f ∈ B θp,p ( Z ) has a ν -a.e. representative which is η -H¨older continuous for any 0 < η < − s ν /p . • If 0 < s ν < < p , then every f ∈ B θp,p ( Z ) has a ν -a.e. representative which is(1 − /p )-H¨older continuous. xtension and trace results for doubling metric measure spaces and their hyperbolic fillings 51 References Adams, D. 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